work done problem solving

Work done by force – problems and solutions

1. A person pulls a block 2 m along a horizontal surface by a constant force F = 20 N. Determine the work done by force F acting on the block.

Angle (θ ) = 0

W = F d = w h = m g h

Weight (w) = m g = (1 kg)(10 m/s 2 ) = 10 kg m/s 2 = 10 N

k = F / x = w / x = m g / x

W = – (250)(0.0004)

Work done by force F :

W net = 20 – 4

Wanted: Work (W)

Force (F) = 200 Newton

W = (200 Newton)(2 meters)

Force (F) = 50 Newton

W = 475 Joule

Displacement (s) = 4 meters

11. Based on figure below, if work done by net force is 375 Joule, determine object’s displacement.

12. The activities below w hich do not do work is …

The equation of work :

Displacement = 0 so work = 0.

Work done by the resistance force of the ground :

9.10 Rate Word Problems: Work and Time

If it takes Felicia 4 hours to paint a room and her daughter Katy 12 hours to paint the same room, then working together, they could paint the room in 3 hours. The equation used to solve problems of this type is one of reciprocals. It is derived as follows:

[latex]\text{rate}\times \text{time}=\text{work done}[/latex]

For this problem:

[latex]\begin{array}{rrrl} \text{Felicia's rate: }&F_{\text{rate}}\times 4 \text{ h}&=&1\text{ room} \\ \\ \text{Katy's rate: }&K_{\text{rate}}\times 12 \text{ h}&=&1\text{ room} \\ \\ \text{Isolating for their rates: }&F&=&\dfrac{1}{4}\text{ h and }K = \dfrac{1}{12}\text{ h} \end{array}[/latex]

To make this into a solvable equation, find the total time [latex](T)[/latex] needed for Felicia and Katy to paint the room. This time is the sum of the rates of Felicia and Katy, or:

[latex]\begin{array}{rcrl} \text{Total time: } &T \left(\dfrac{1}{4}\text{ h}+\dfrac{1}{12}\text{ h}\right)&=&1\text{ room} \\ \\ \text{This can also be written as: }&\dfrac{1}{4}\text{ h}+\dfrac{1}{12}\text{ h}&=&\dfrac{1 \text{ room}}{T} \\ \\ \text{Solving this yields:}&0.25+0.083&=&\dfrac{1 \text{ room}}{T} \\ \\ &0.333&=&\dfrac{1 \text{ room}}{T} \\ \\ &t&=&\dfrac{1}{0.333}\text{ or }\dfrac{3\text{ h}}{\text{room}} \end{array}[/latex]

Example 9.10.1

Karl can clean a room in 3 hours. If his little sister Kyra helps, they can clean it in 2.4 hours. How long would it take Kyra to do the job alone?

The equation to solve is:

[latex]\begin{array}{rrrrl} \dfrac{1}{3}\text{ h}&+&\dfrac{1}{K}&=&\dfrac{1}{2.4}\text{ h} \\ \\ &&\dfrac{1}{K}&=&\dfrac{1}{2.4}\text{ h}-\dfrac{1}{3}\text{ h}\\ \\ &&\dfrac{1}{K}&=&0.0833\text{ or }K=12\text{ h} \end{array}[/latex]

Example 9.10.2

Doug takes twice as long as Becky to complete a project. Together they can complete the project in 10 hours. How long will it take each of them to complete the project alone?

[latex]\begin{array}{rrl} \dfrac{1}{R}+\dfrac{1}{2R}&=&\dfrac{1}{10}\text{ h,} \\ \text{where Doug's rate (} \dfrac{1}{D}\text{)}& =& \dfrac{1}{2}\times \text{ Becky's (}\dfrac{1}{R}\text{) rate.} \\ \\ \text{Sum the rates: }\dfrac{1}{R}+\dfrac{1}{2R}&=&\dfrac{2}{2R} + \dfrac{1}{2R} = \dfrac{3}{2R} \\ \\ \text{Solve for R: }\dfrac{3}{2R}&=&\dfrac{1}{10}\text{ h} \\ \text{which means }\dfrac{1}{R}&=&\dfrac{1}{10}\times\dfrac{2}{3}\text{ h} \\ \text{so }\dfrac{1}{R}& =& \dfrac{2}{30} \\ \text{ or }R &= &\dfrac{30}{2} \end{array}[/latex]

This means that the time it takes Becky to complete the project alone is [latex]15\text{ h}[/latex].

Since it takes Doug twice as long as Becky, the time for Doug is [latex]30\text{ h}[/latex].

Example 9.10.3

Joey can build a large shed in 10 days less than Cosmo can. If they built it together, it would take them 12 days. How long would it take each of them working alone?

[latex]\begin{array}{rl} \text{The equation to solve:}& \dfrac{1}{(C-10)}+\dfrac{1}{C}=\dfrac{1}{12}, \text{ where }J=C-10 \\ \\ \text{Multiply each term by the LCD:}&(C-10)(C)(12) \\ \\ \text{This leaves}&12C+12(C-10)=C(C-10) \\ \\ \text{Multiplying this out:}&12C+12C-120=C^2-10C \\ \\ \text{Which simplifies to}&C^2-34C+120=0 \\ \\ \text{Which will factor to}& (C-30)(C-4) = 0 \end{array}[/latex]

Cosmo can build the large shed in either 30 days or 4 days. Joey, therefore, can build the shed in 20 days or −6 days (rejected).

The solution is Cosmo takes 30 days to build and Joey takes 20 days.

Example 9.10.4

Clark can complete a job in one hour less than his apprentice. Together, they do the job in 1 hour and 12 minutes. How long would it take each of them working alone?

[latex]\begin{array}{rl} \text{Convert everything to hours:} & 1\text{ h }12\text{ min}=\dfrac{72}{60} \text{ h}=\dfrac{6}{5}\text{ h}\\ \\ \text{The equation to solve is} & \dfrac{1}{A}+\dfrac{1}{A-1}=\dfrac{1}{\dfrac{6}{5}}=\dfrac{5}{6}\\ \\ \text{Therefore the equation is} & \dfrac{1}{A}+\dfrac{1}{A-1}=\dfrac{5}{6} \\ \\ \begin{array}{r} \text{To remove the fractions, } \\ \text{multiply each term by the LCD} \end{array} & (A)(A-1)(6)\\ \\ \text{This leaves} & 6(A)+6(A-1)=5(A)(A-1) \\ \\ \text{Multiplying this out gives} & 6A-6+6A=5A^2-5A \\ \\ \text{Which simplifies to} & 5A^2-17A +6=0 \\ \\ \text{This will factor to} & (5A-2)(A-3)=0 \end{array}[/latex]

The apprentice can do the job in either [latex]\dfrac{2}{5}[/latex] h (reject) or 3 h. Clark takes 2 h.

Example 9.10.5

A sink can be filled by a pipe in 5 minutes, but it takes 7 minutes to drain a full sink. If both the pipe and the drain are open, how long will it take to fill the sink?

The 7 minutes to drain will be subtracted.

[latex]\begin{array}{rl} \text{The equation to solve is} & \dfrac{1}{5}-\dfrac{1}{7}=\dfrac{1}{X} \\ \\ \begin{array}{r} \text{To remove the fractions,} \\ \text{multiply each term by the LCD}\end{array} & (5)(7)(X)\\ \\ \text{This leaves } & (7)(X)-(5)(X)=(5)(7)\\ \\ \text{Multiplying this out gives} & 7X-5X=35\\ \\ \text{Which simplifies to} & 2X=35\text{ or }X=\dfrac{35}{2}\text{ or }17.5 \end{array}[/latex]

17.5 min or 17 min 30 sec is the solution

For Questions 1 to 8, write the formula defining the relation. Do Not Solve!!

Bill’s father can paint a room in 2 hours less than it would take Bill to paint it. Working together, they can complete the job in 2 hours and 24 minutes. How much time would each require working alone?
Of two inlet pipes, the smaller pipe takes four hours longer than the larger pipe to fill a pool. When both pipes are open, the pool is filled in three hours and forty-five minutes. If only the larger pipe is open, how many hours are required to fill the pool?
Jack can wash and wax the family car in one hour less than it would take Bob. The two working together can complete the job in 1.2 hours. How much time would each require if they worked alone?
If Yousef can do a piece of work alone in 6 days, and Bridgit can do it alone in 4 days, how long will it take the two to complete the job working together?
Working alone, it takes John 8 hours longer than Carlos to do a job. Working together, they can do the job in 3 hours. How long would it take each to do the job working alone?
Working alone, Maryam can do a piece of work in 3 days that Noor can do in 4 days and Elana can do in 5 days. How long will it take them to do it working together?
Raj can do a piece of work in 4 days and Rubi can do it in half the time. How long would it take them to do the work together?
A cistern can be filled by one pipe in 20 minutes and by another in 30 minutes. How long would it take both pipes together to fill the tank?

For Questions 9 to 20, find and solve the equation describing the relationship.

If an apprentice can do a piece of work in 24 days, and apprentice and instructor together can do it in 6 days, how long would it take the instructor to do the work alone?
A carpenter and his assistant can do a piece of work in 3.75 days. If the carpenter himself could do the work alone in 5 days, how long would the assistant take to do the work alone?
If Sam can do a certain job in 3 days, while it would take Fred 6 days to do the same job, how long would it take them, working together, to complete the job?
Tim can finish a certain job in 10 hours. It takes his wife JoAnn only 8 hours to do the same job. If they work together, how long will it take them to complete the job?
Two people working together can complete a job in 6 hours. If one of them works twice as fast as the other, how long would it take the slower person, working alone, to do the job?
If two people working together can do a job in 3 hours, how long would it take the faster person to do the same job if one of them is 3 times as fast as the other?
A water tank can be filled by an inlet pipe in 8 hours. It takes twice that long for the outlet pipe to empty the tank. How long would it take to fill the tank if both pipes were open?
A sink can be filled from the faucet in 5 minutes. It takes only 3 minutes to empty the sink when the drain is open. If the sink is full and both the faucet and the drain are open, how long will it take to empty the sink?
It takes 10 hours to fill a pool with the inlet pipe. It can be emptied in 15 hours with the outlet pipe. If the pool is half full to begin with, how long will it take to fill it from there if both pipes are open?
A sink is ¼ full when both the faucet and the drain are opened. The faucet alone can fill the sink in 6 minutes, while it takes 8 minutes to empty it with the drain. How long will it take to fill the remaining ¾ of the sink?
A sink has two faucets: one for hot water and one for cold water. The sink can be filled by a cold-water faucet in 3.5 minutes. If both faucets are open, the sink is filled in 2.1 minutes. How long does it take to fill the sink with just the hot-water faucet open?
A water tank is being filled by two inlet pipes. Pipe A can fill the tank in 4.5 hours, while both pipes together can fill the tank in 2 hours. How long does it take to fill the tank using only pipe B?

Answer Key 9.10

Share This Book

HW Guidelines
Study Skills Quiz
Find Local Tutors
Demo MathHelp.com
Join MathHelp.com

Select a Course Below

ACCUPLACER Math
Math Placement Test
PRAXIS Math
+ more tests
5th Grade Math
6th Grade Math
Pre-Algebra
College Pre-Algebra
Introductory Algebra
Intermediate Algebra
College Algebra

"Work" Word Problems

Painting & Pipes Tubs & Man-Hours Unequal Times Etc.

"Work" problems usually involve situations such as two people working together to paint a house. You are usually told how long each person takes to paint a similarly-sized house, and you are asked how long it will take the two of them to paint the house when they work together.

Many of these problems are not terribly realistic — since when can two laser printers work together on printing one report? — but it's the technique that they want you to learn, not the applicability to "real life".

The method of solution for "work" problems is not obvious, so don't feel bad if you're totally lost at the moment. There is a "trick" to doing work problems: you have to think of the problem in terms of how much each person / machine / whatever does in a given unit of time . For instance:

Content Continues Below

MathHelp.com

Suppose one painter can paint the entire house in twelve hours, and the second painter takes eight hours to paint a similarly-sized house. How long would it take the two painters together to paint the house?

To find out how much they can do together per hour , I make the necessary assumption that their labors are additive (in other words, that they never get in each other's way in any manner), and I add together what they can do individually per hour . So, per hour, their labors are:

But the exercise didn't ask me how much they can do per hour; it asked me how long they'll take to finish one whole job, working togets. So now I'll pick the variable " t " to stand for how long they take (that is, the time they take) to do the job together. Then they can do:

This gives me an expression for their combined hourly rate. I already had a numerical expression for their combined hourly rate. So, setting these two expressions equal, I get:

I can solve by flipping the equation; I get:

An hour has sixty minutes, so 0.8 of an hour has forty-eight minutes. Then:

They can complete the job together in 4 hours and 48 minutes.

The important thing to understand about the above example is that the key was in converting how long each person took to complete the task into a rate.

hours to complete job:

first painter: 12

second painter: 8

together: t

Since the unit for completion was "hours", I converted each time to an hourly rate; that is, I restated everything in terms of how much of the entire task could be completed per hour. To do this, I simply inverted each value for "hours to complete job":

completed per hour:

Then, assuming that their per-hour rates were additive, I added the portion that each could do per hour, summed them, and set this equal to the "together" rate:

adding their labor:

As you can see in the above example, "work" problems commonly create rational equations . But the equations themselves are usually pretty simple to solve.

One pipe can fill a pool 1.25 times as fast as a second pipe. When both pipes are opened, they fill the pool in five hours. How long would it take to fill the pool if only the slower pipe is used?

My first step is to list the times taken by each pipe to fill the pool, and how long the two pipes take together. In this case, I know the "together" time, but not the individual times. One of the pipes' times is expressed in terms of the other pipe's time, so I'll pick a variable to stand for one of these times.

Since the faster pipe's time to completion is defined in terms of the second pipe's time, I'll pick a variable for the slower pipe's time, and then use this to create an expression for the faster pipe's time:

slow pipe: s

together: 5

Next, I'll convert all of the completion times to per-hour rates:

Then I make the necessary assumption that the pipes' contributions are additive (which is reasonable, in this case), add the two pipes' contributions, and set this equal to the combined per-hour rate:

multiplying through by 20 s (being the lowest common denominator of all the fractional terms):

20 + 25 = 4 s

45/4 = 11.25 = s

They asked me for the time of the slower pipe, so I don't need to find the time for the faster pipe. My answer is:

The slower pipe takes 11.25 hours.

Note: I could have picked a variable for the faster pipe, and then defined the time for the slower pipe in terms of this variable. If you're not sure how you'd do this, then think about it in terms of nicer numbers: If someone goes twice as fast as you, then you take twice as long as he does; if he goes three times as fast as you, then you take three times as long as him. In this case, if he goes 1.25 times as fast, then you take 1.25 times as long. So the variables could have been " f " for the number of hours the faster pipe takes, and then the number of hours for the slower pipe would have been " 1.25 f ".

URL: https://www.purplemath.com/modules/workprob.htm

Page 1 Page 2 Page 3 Page 4

Standardized Test Prep

College math, homeschool math, share this page.

Terms of Use
About Purplemath
About the Author
Tutoring from PM
Advertising
Linking to PM
Site licencing

Visit Our Profiles

Learning Objectives

By the end of this section, you will be able to:

Represent the work done by any force
Evaluate the work done for various forces

In physics, work is done on an object when energy is transferred to the object. In other words, work is done when a force acts on something that undergoes a displacement from one position to another. Forces can vary as a function of position, and displacements can be along various paths between two points. We first define the increment of work dW done by a force F → F → acting through an infinitesimal displacement d r → d r → as the dot product of these two vectors:

Then, we can add up the contributions for infinitesimal displacements, along a path between two positions, to get the total work.

Work Done by a Force

The work done by a force is the integral of the force with respect to displacement along the path of the displacement:

The vectors involved in the definition of the work done by a force acting on a particle are illustrated in Figure 7.2 . While in general, Equation 7.2 requires mathematics beyond the scope of this text, in many simple situations this integral becomes a familiar integral in one variable. We will examine several such examples and restrict our discussion to these cases.

We choose to express the dot product in terms of the magnitudes of the vectors and the cosine of the angle between them, because the meaning of the dot product for work can be put into words more directly in terms of magnitudes and angles. We could equally well have expressed the dot product in terms of the various components introduced in Vectors . In two dimensions, these were the x - and y -components in Cartesian coordinates, or the r - and φ φ -components in polar coordinates; in three dimensions, it was just x -, y -, and z -components. Which choice is more convenient depends on the situation. In words, you can express Equation 7.1 for the work done by a force acting over a displacement as a product of one component acting parallel to the other component. From the properties of vectors, it doesn’t matter if you take the component of the force parallel to the displacement or the component of the displacement parallel to the force—you get the same result either way.

Recall that the magnitude of a force times the cosine of the angle the force makes with a given direction is the component of the force in the given direction. The components of a vector can be positive, negative, or zero, depending on whether the angle between the vector and the component-direction is between 0 ° 0 ° and 90 ° 90 ° or 90 ° 90 ° and 180 ° 180 ° , or is equal to 90 ° 90 ° . As a result, the work done by a force can be positive, negative, or zero, depending on whether the force is generally in the direction of the displacement, generally opposite to the displacement, or perpendicular to the displacement. The maximum work is done by a given force when it is along the direction of the displacement ( cos θ = ± 1 cos θ = ± 1 ), and zero work is done when the force is perpendicular to the displacement ( cos θ = 0 cos θ = 0 ).

The units of work are units of force multiplied by units of length, which in the SI system is newtons times meters, N · m. N · m. This combination is called a joule , for historical reasons that we will mention later, and is abbreviated as J. In the English system, still used in the United States, the unit of force is the pound (lb) and the unit of distance is the foot (ft), so the unit of work is the foot-pound ( ft · lb ) . ( ft · lb ) .

Work Done by Constant Forces and Contact Forces

The simplest work to evaluate is that done by a force that is constant in magnitude and direction. In this case, we can factor out the force; the remaining integral is just the total displacement, which only depends on the end points A and B , but not on the path between them:

Figure 7.3 (a) shows a person exerting a constant force F → F → along the handle of a lawn mower, which makes an angle θ θ with the horizontal. The horizontal displacement of the lawn mower, over which the force acts, is d → . d → . The work done on the lawn mower is W = F → · d → = F d cos θ W = F → · d → = F d cos θ , which the figure also illustrates as the horizontal component of the force times the magnitude of the displacement.

Figure 7.3 (b) shows a person holding a briefcase. The person must exert an upward force, equal in magnitude to the weight of the briefcase, but this force does no work, because the displacement over which it acts is zero.

In Figure 7.3 (c), where the person in (b) is walking horizontally with constant speed, the work done by the person on the briefcase is still zero, but now because the angle between the force exerted and the displacement is 90 ° 90 ° ( F → F → perpendicular to d → d → ) and cos 90 ° = 0 cos 90 ° = 0 .

Example 7.1

Calculating the work you do to push a lawn mower.

Substituting the known values gives

Significance

When you mow the grass, other forces act on the lawn mower besides the force you exert—namely, the contact force of the ground and the gravitational force of Earth. Let’s consider the work done by these forces in general. For an object moving on a surface, the displacement d r → d r → is tangent to the surface. The part of the contact force on the object that is perpendicular to the surface is the normal force N → . N → . Since the cosine of the angle between the normal and the tangent to a surface is zero, we have

The normal force never does work under these circumstances. (Note that if the displacement d r → d r → did have a relative component perpendicular to the surface, the object would either leave the surface or break through it, and there would no longer be any normal contact force. However, if the object is more than a particle, and has an internal structure, the normal contact force can do work on it, for example, by displacing it or deforming its shape. This will be mentioned in the next chapter.)

The part of the contact force on the object that is parallel to the surface is friction, f → . f → . For this object sliding along the surface, kinetic friction f → k f → k is opposite to d r → , d r → , relative to the surface, so the work done by kinetic friction is negative. If the magnitude of f → k f → k is constant (as it would be if all the other forces on the object were constant), then the work done by friction is

where | l A B | | l A B | is the path length on the surface. The force of static friction does no work in the reference frame between two surfaces because there is never displacement between the surfaces. As an external force, static friction can do work. Static friction can keep someone from sliding off a sled when the sled is moving and perform positive work on the person. If you’re driving your car at the speed limit on a straight, level stretch of highway, the negative work done by air resistance is balanced by the positive work done by the static friction of the road on the drive wheels. You can pull the rug out from under an object in such a way that it slides backward relative to the rug, but forward relative to the floor. In this case, kinetic friction exerted by the rug on the object could be in the same direction as the displacement of the object, relative to the floor, and do positive work. The bottom line is that you need to analyze each particular case to determine the work done by the forces, whether positive, negative or zero.

Example 7.2

Moving a couch.

The work done by friction i W = − ( 0.6 ) ( 1 kN ) ( 3 m + 1 m ) = − 2.4 kJ . W = − ( 0.6 ) ( 1 kN ) ( 3 m + 1 m ) = − 2.4 kJ .
The length of the path along the hypotenuse is 10 m 10 m , so the total work done against friction is W = ( 0.6 ) ( 1 kN ) ( 3 m + 1 m + 1 0 m ) = 4.3 kJ . W = ( 0.6 ) ( 1 kN ) ( 3 m + 1 m + 1 0 m ) = 4.3 kJ .

Check Your Understanding 7.1

Can kinetic friction ever be a constant force for all paths?

The other force on the lawn mower mentioned above was Earth’s gravitational force, or the weight of the mower. Near the surface of Earth, the gravitational force on an object of mass m has a constant magnitude, mg , and constant direction, vertically down. Therefore, the work done by gravity on an object is the dot product of its weight and its displacement. In many cases, it is convenient to express the dot product for gravitational work in terms of the x -, y -, and z -components of the vectors. A typical coordinate system has the x -axis horizontal and the y -axis vertically up. Then the gravitational force is − m g j ^ , − m g j ^ , so the work done by gravity, over any path from A to B , is

The work done by a constant force of gravity on an object depends only on the object’s weight and the difference in height through which the object is displaced. Gravity does negative work on an object that moves upward ( y B > y A y B > y A ), or, in other words, you must do positive work against gravity to lift an object upward. Alternately, gravity does positive work on an object that moves downward ( y B < y A y B < y A ), or you do negative work against gravity to “lift” an object downward, controlling its descent so it doesn’t drop to the ground. (“Lift” is used as opposed to “drop”.)

Example 7.3

Shelving a book.

Since the book starts on the shelf and is lifted down y B − y A = − 1 m y B − y A = − 1 m , we have W = − ( 20 N ) ( − 1 m ) = 20 J . W = − ( 20 N ) ( − 1 m ) = 20 J .
There is zero difference in height for any path that begins and ends at the same place on the shelf, so W = 0 . W = 0 .

Check Your Understanding 7.2

Can Earth’s gravity ever be a constant force for all paths?

Work Done by Forces that Vary

In general, forces may vary in magnitude and direction at points in space, and paths between two points may be curved. The infinitesimal work done by a variable force can be expressed in terms of the components of the force and the displacement along the path,

Here, the components of the force are functions of position along the path, and the displacements depend on the equations of the path. (Although we chose to illustrate dW in Cartesian coordinates, other coordinates are better suited to some situations.) Equation 7.2 defines the total work as a line integral, or the limit of a sum of infinitesimal amounts of work. The physical concept of work is straightforward: you calculate the work for tiny displacements and add them up. Sometimes the mathematics can seem complicated, but the following example demonstrates how cleanly they can operate.

Example 7.4

Work done by a variable force over a curved path.

Then, the integral for the work is just a definite integral of a function of x .

The integral of x 2 x 2 is x 3 / 3 , x 3 / 3 , so

Check Your Understanding 7.3

Find the work done by the same force in Example 7.4 over a cubic path, y = ( 0.25 m −2 ) x 3 y = ( 0.25 m −2 ) x 3 , between the same points A = ( 0 , 0 ) A = ( 0 , 0 ) and B = ( 2 m, 2 m ) . B = ( 2 m, 2 m ) .

One very important and widely applicable variable force is the force exerted by a perfectly elastic spring, which satisfies Hooke’s law F → = − k Δ x → , F → = − k Δ x → , where k is the spring constant, and Δ x → = x → − x → eq Δ x → = x → − x → eq is the displacement from the spring’s unstretched (equilibrium) position ( Newton’s Laws of Motion ). Note that the unstretched position is only the same as the equilibrium position if no other forces are acting (or, if they are, they cancel one another). Forces between molecules, or in any system undergoing small displacements from a stable equilibrium, behave approximately like a spring force.

To calculate the work done by a spring force, we can choose the x -axis along the length of the spring, in the direction of increasing length, as in Figure 7.7 , with the origin at the equilibrium position x eq = 0 . x eq = 0 . (Then positive x corresponds to a stretch and negative x to a compression.) With this choice of coordinates, the spring force has only an x -component, F x = − k x F x = − k x , and the work done when x changes from x A x A to x B x B is

Notice that W A B W A B depends only on the starting and ending points, A and B , and is independent of the actual path between them, as long as it starts at A and ends at B. That is, the actual path could involve going back and forth before ending.

Another interesting thing to notice about Equation 7.5 is that, for this one-dimensional case, you can readily see the correspondence between the work done by a force and the area under the curve of the force versus its displacement. Recall that, in general, a one-dimensional integral is the limit of the sum of infinitesimals, f ( x ) d x f ( x ) d x , representing the area of strips, as shown in Figure 7.8 . In Equation 7.5 , since F = − k x F = − k x is a straight line with slope − k − k , when plotted versus x , the “area” under the line is just an algebraic combination of triangular “areas,” where “areas” above the x -axis are positive and those below are negative, as shown in Figure 7.9 . The magnitude of one of these “areas” is just one-half the triangle’s base, along the x -axis, times the triangle’s height, along the force axis. (There are quotation marks around “area” because this base-height product has the units of work, rather than square meters.)

Example 7.5

Work done by a spring force.

For part (a), x A = 0 x A = 0 and x B = 6 cm x B = 6 cm ; for part (b), x B = 6 cm x B = 6 cm and x B = 12 cm x B = 12 cm . In part (a), the work is given and you can solve for the spring constant; in part (b), you can use the value of k , from part (a), to solve for the work.

W = 0.54 J = 1 2 k [ ( 6 cm ) 2 − 0 ] W = 0.54 J = 1 2 k [ ( 6 cm ) 2 − 0 ] , so k = 3 N/cm . k = 3 N/cm .
W = 1 2 ( 3 N/cm ) [ ( 12 cm ) 2 − ( 6 cm ) 2 ] = 1.62 J . W = 1 2 ( 3 N/cm ) [ ( 12 cm ) 2 − ( 6 cm ) 2 ] = 1.62 J .

Check Your Understanding 7.4

The spring in Example 7.5 is compressed 6 cm from its equilibrium length. (a) Does the spring force do positive or negative work and (b) what is the magnitude?

This book may not be used in the training of large language models or otherwise be ingested into large language models or generative AI offerings without OpenStax's permission.

Want to cite, share, or modify this book? This book uses the Creative Commons Attribution License and you must attribute OpenStax.

Access for free at https://openstax.org/books/university-physics-volume-1/pages/1-introduction

Authors: William Moebs, Samuel J. Ling, Jeff Sanny
Publisher/website: OpenStax
Book title: University Physics Volume 1
Publication date: Sep 19, 2016
Location: Houston, Texas
Book URL: https://openstax.org/books/university-physics-volume-1/pages/1-introduction
Section URL: https://openstax.org/books/university-physics-volume-1/pages/7-1-work

© Jul 23, 2024 OpenStax. Textbook content produced by OpenStax is licensed under a Creative Commons Attribution License . The OpenStax name, OpenStax logo, OpenStax book covers, OpenStax CNX name, and OpenStax CNX logo are not subject to the Creative Commons license and may not be reproduced without the prior and express written consent of Rice University.

Work Rate Problems with Solutions

A set of problems related to work and rate of work is presented with detailed solutions.

It takes 1.5 hours for Tim to mow the lawn. Linda can mow the same lawn in 2 hours. How long will it take John and Linda, work together, to mow the lawn?

John: 1 / 1.5 and Linda 1 / 2

It takes 6 hours for pump A, used alone, to fill a tank of water. Pump B used alone takes 8 hours to fill the same tank. We want to use three pumps: A, B and another pump C to fill the tank in 2 hours. What should be the rate of pump C? How long would it take pump C, used alone, to fill the tank?

A: 1 / 6 and B: 1 / 8

R = 1 / 4.8 , rate of pump C.

t × (1 / 4.8) = 1

t = 4.8 hours , the time it takes pump C to fill the tank.

A tank can be filled by pipe A in 5 hours and by pipe B in 8 hours, each pump working on its own. When the tank is full and a drainage hole is open, the water is drained in 20 hours. If initially the tank was empty and someone started the two pumps together but left the drainage hole open, how long does it take for the tank to be filled?

pump A: 1 / 5 , pump B: 1 / 8 , drainage hole: 1 / 20
water into the tank however the drainage hole water out of the tank, hence
t ( 1 / 5 + 1 / 8 - 1 / 20) = 1

t = 3.6 hours.

A swimming pool can be filled by pipe A in 3 hours and by pipe B in 6 hours, each pump working on its own. At 9 am pump A is started. At what time will the swimming pool be filled if pump B is started at 10 am?

pump A: 1 / 3 , pump B: 1 / 6

t = 7 / 3 hours = 2.3 hours = 2 hours 20 minutes.

9 + 2:20 = 11:20

How to Master Work Problems: A Comprehensive Step-by-Step Guide

Understanding work problems in mathematics often involves dealing with scenarios where different people (or machines) contribute to completing a task. These problems can be solved by using the formula $W=R×T$, where $W$ is Work, $R$ is Rate, and $T$ is Time. Here is a step-by-step guide to help you understand and solve these problems:

Step-by-step Guide to Master Work Problems

Step 1: understand the problem.

Read the Problem Carefully: Identify what the problem is asking. Pay attention to the time taken by each person or machine to complete the work independently.
Define the Work: Usually, the total work done is considered as $1$ unit. For example, mowing a lawn, painting a wall, or filling a tank is each considered as 1 unit of work.

Step 2: Determine the Rates

Calculate Individual Rates: Determine the rate at which each person or machine completes the work. If a person completes the work in $T$ hours, their rate is $\frac{1}{T}$ units of work per hour.
Combine Rates for Collaborative Work: If multiple people or machines work together, add their rates. For instance, if Person $A$ has a rate of $\frac{1}{T_{A}}$ and Person $B$ has a rate of $\frac{1}{T_{B}}$, their combined rate is$\frac{1}{T_{A}}+\frac{1}{T_{B}}$ .

Step 3: Set Up the Equation

Use the Work Formula: The formula $W=R×T$ is pivotal. For combined work, set $W=1$ and use the combined rate for $R$.
Formulate the Equation: The equation usually looks like $1=(\frac{1}{T_{A}}+\frac{1}{T_{B}})×T$, where $T$ is the time taken for the combined work.

Step 4: Solve for the Unknown

Rearrange the Equation: Isolate the variable you are solving for. This might involve algebraic manipulation.
Solve Mathematically: Use algebra to find the value of the unknown variable. This might involve finding a common denominator, simplifying fractions, or solving linear equations.

Step 5: Check Your Solution

Verify the Answer: Plug your answer back into the equation to see if it makes sense. Check if the units and the context align correctly.
Consider Practical Implications: Ensure that the solution is practical and makes sense in the context of the problem.

Understanding each step and applying it to various problems will enhance your ability to tackle work problems effectively.

Sarah can clean a room in $2$ hours. When she works with her friend Lisa, they can clean it in $1.5$ hours. How long would it take Lisa to clean the room alone?

Sarah’s rate $=\frac{1}{2}$ room per hour.
Together, their rate $=\frac{1}{1.5}$ room per hour.
Let Lisa’s time be $T$ hours, so her rate $=\frac{1}{T}$.

Combine the rates: $\frac{1}{2}+\frac{1}{T}=\frac{1}{1.5}$

Solve for $T$: $\frac{1}{T}=\frac{2}{3}-\frac{1}{2}=\frac{4-3}{6}=\frac{1}{6}$ So, $T=6$ hours.

Kevin can type a document in $5$ hours. Working together with Rachel, they can type it in $4$ hours. How long would it take Rachel alone?

Kevin’s rate $=\frac{1}{5}$ document per hour.
Together, their rate $=\frac{1}{4}$ document per hour.
Rachel’s rate $=\frac{1}{T}$.

Combine the rates: $\frac{1}{5}+\frac{1}{T}=\frac{1}{4}$

Solve for $T$: $\frac{1}{T}=\frac{1}{4}-\frac{1}{5}=\frac{5-4}{20}=\frac{1}{20}$ So, $T=20$ hours.

by: Effortless Math Team about 10 months ago (category: Articles )

Effortless Math Team

Related to this article, more math articles.

Evaluating Variables and Expressions
How to Find Magnitude of Vectors?
6th Grade STAAR Math Worksheets: FREE & Printable
How to Find the Perimeter of Right-Angled Triangle?
PSAT 10 Math Formulas
How to Prepare for the Pre-Algebra Test?
How to Solve Word Problems Involving One-step Inequalities?
How to Solve Linear Equations in Two Variables?
Full-Length TASC Math Practice Test-Answers and Explanations
How to Grasp Parallel Vectors

What people say about "How to Master Work Problems: A Comprehensive Step-by-Step Guide - Effortless Math: We Help Students Learn to LOVE Mathematics"?

No one replied yet.

M-STEP Grade 6 Math for Beginners The Ultimate Step by Step Guide to Preparing for the M-STEP Math Test

Algebra 2 workbook a comprehensive review and step-by-step guide for mastering essential math skills, dat quantitative reasoning workbook 2018 – 2019 a comprehensive review and step-by-step guide to preparing for the dat math, algebra 1 workbook a comprehensive review and step-by-step guide for mastering essential math skills, gre mathematics workbook 2018-2019 a comprehensive review and step-by-step guide to preparing for the gre math, act mathematics workbook 2018 – 2019 a comprehensive review and step-by-step guide to preparing for the act math, tsi mathematics workbook 2018 – 2019 a comprehensive review and step-by-step guide to preparing for the tsi math, grade 7 georgia milestones mathematics workbook 2018-2019 a comprehensive review and step-by-step guide to preparing for the gmas math test, grade 6 georgia milestones mathematics workbook 2018-2019 a comprehensive review and step-by-step guide to preparing for the gmas math test, grade 5 georgia milestones mathematics workbook 2018-2019 a comprehensive review and step-by-step guide to preparing for the gmas math test, grade 5 fsa mathematics workbook 2018 – 2019 a comprehensive review and step-by-step guide to preparing for the fsa math test, grade 8 fsa mathematics workbook 2018-2019 a comprehensive review and step-by-step guide to preparing for the fsa math test, grade 7 fsa mathematics workbook 2018 – 2019 a comprehensive review and step-by-step guide to preparing for the fsa math test, grade 6 fsa mathematics workbook 2018 – 2019 a comprehensive review and step-by-step guide to preparing for the fsa math test.

ATI TEAS 6 Math
ISEE Upper Level Math
SSAT Upper-Level Math
Praxis Core Math
8th Grade STAAR Math

Limited time only!

Save Over 45 %

It was $89.99 now it is $49.99

Login and use all of our services.

Effortless Math services are waiting for you. login faster!

Register Fast!

Password will be generated automatically and sent to your email.

After registration you can change your password if you want.

Math Worksheets
Math Courses
Math Topics
Math Puzzles
Math eBooks
GED Math Books
HiSET Math Books
ACT Math Books
ISEE Math Books
ACCUPLACER Books
Premium Membership
Youtube Videos

Effortless Math provides unofficial test prep products for a variety of tests and exams. All trademarks are property of their respective trademark owners.

Bulk Orders
Refund Policy

Exam Center
Ticket Center
Flash Cards
Work and Energy

Physics Work Problems for High Schools

In this tutorial, we want to practice some problems on work in physics. All these questions are easy and helpful for your high school homework.

Work Problems: Constant Force

Problem (1): A constant force of 1200 N is required to push a car along a straight line. A person displaces the car by 45 m. How much work is done by the person?

Solution : If a constant force $F$ acts on an object over a distance of $d$, and $F$ is parallel to $d$, then the work done by force $F$ is the product of the force times distance.

In this case, a force of $1200\,{\rm N}$ displaces the car $45\,{\rm m}$. The pushing force is parallel to the displacement. So, the work done by the person is equal to \[W=Fd=1200\times 45=54000\,{\rm J}\] The SI unit of work is the joule, ${\rm J}$.

Problem (2): You lift a book of mass $2\,{\rm kg}$ at constant speed straight upward a distance of $2\,{\rm m}$. How much work is done during this lifting by you?

Solution : The force you apply to lift the book must be balanced with the book's weight. So, the exerted force on the book is \[F=mg=2\times 10=20\quad{\rm N}\] The book is lifted 2 meters vertically. The force and displacement are both parallel to each other, so the work done by the person is the product of them. \[W=Fd=20\times 2=40\quad {\rm J}\]

Problem (3): A force of $F=20\,{\rm N}$ at an angle of $37^\circ$ is applied to a 3-kg object initially at rest. The object has displaced a distance of $25\,{\rm m}$ over a frictionless horizontal table. Determine the work done by (a) The applied force (b) The normal force exerted by the table (c) The force of gravity

Solution : In this problem, the force makes an angle with the displacement. In such cases, we should use the work formula $W=Fd\cos\theta$ where $\theta$ is the angle between force $F$ and displacement. To this object, an external force $F$, normal force $F_N$, and a gravity force $w=mg$ are applied.

(a) Using vector decompositions, the component of the force parallel to the displacement is found to be $F_{\parallel}=F\cos \theta$. Thus, the product of this component parallel to the displacement times the magnitude of displacement gives us the work done by external force $F$ as below \begin{align*} W_F&=\underbrace{F\cos\theta}_{F_{\parallel}}d\\\\ &=(20\times \cos 37^\circ)(25)\\\\&=400\quad {\rm J}\end{align*} (b) Now, we want to find the work done by the normal force. But let's define what the normal force is.

In physics, ''normal'' means perpendicular. When an object is in contact with a surface, a contact force is exerted on the object. The component of the contact force perpendicular to the surface is called the normal force.

Thus, by definition, the normal force is always perpendicular to the displacement. So, the angle between $F_N$ and displacement $d$ is $90^\circ$. Hence, the work done by the normal force is determined to be \[W_N=F_N d\cos\theta=(30)(25)\cos 90^\circ=0\] (c) The weight of the object is the same as the force of gravity. This force applies to the object vertically downward, and the displacement of the object is horizontal. So, again, the angle between these two vectors is $\theta=90^\circ$. Hence, the work done by the force of gravity is zero.

Problem (4): A person pulls a crate using a force of $56\,{\rm N}$ which makes an angle of $25^\circ$ with the horizontal. The floor is frictionless. How much work does he do in pulling the crate over a horizontal distance of $200\,{\rm m}$?

Solution : The component of the external force parallel to the displacement does work on an object over a distance of $d$. In all work problems in physics, this force component parallel to the displacement is found by the formula $F_{\parallel}=F\cos \theta$. Thus, the work done by this force is computed as below \begin{align*} W&=F_{\parallel}d\\&=(F\cos\theta)d\\&=(56\cos 25^\circ)(200) \\&=10080\quad {\rm J}\end{align*} We could use the work formula from the beginning $W=Fd\cos\theta$ where $\theta$ is the angle between $F$ and $d$.

Problem (5): A worker pushes a cart with a force of $45\,{\rm N}$ directed at an angle of $32^\circ$ below the horizontal. The cart moves at a constant speed. (a) Find the work done by the worker as the cart moves a straight distance of $50\,{\rm m}$. (b) What is the net work done on the cart?

Solution :(a) All information to find the work done by the worker is given, so we have \begin{align*} W&=Fd\cos\theta\\&=(45)(50) \cos 32^\circ\\&=1912.5\quad {\rm J}\end{align*} (b) ''net'' means "total". In all work problems in physics, there are two equivalent methods to find the net work. Identify all forces that are applied to the cart, find their resultant force, and then compute the work done by this net force over a specific distance.

Or compute all works done on the object across a distance individually, then sum them algebraically.

Usually, the second method is easier. We take this approach here.

The cart moves in a straight horizontal path. All forces apply on it are, the worker force $F$, the normal force $F_N$, and the force of gravity or its weight $F_g=mg$. The work done by normal and gravity forces in a horizontal displacement is always zero since the angle between these forces and the displacement is $90^\circ$. So, $W_N=W_g=0$. Hence, the net (total) work done on the object is \[W_{total}=W_N+W_g+W_F=1912.5\,{\rm J}\]

Problem (6): A $1200-{\rm kg}$ box is at rest on a rough floor. How much work is required to move it $5\,{\rm m}$ at a constant speed (a) along the floor against a $230\,{\rm N}$ friction force, (b) vertically?

Solution : In this problem, we want to displace a box $5\,{\rm m}$ horizontally and vertically. In the horizontal direction, there is also kinetic friction.

(a) At constant speed , means there is no acceleration in the course of displacement, so according to Newton's second law $\Sigma F=ma$, the net force on the box must be zero. To meet this condition, the external force $F_p$ applied by a person must cancel out the friction force $f_k$. So, \[F_p=f_k=230\quad {\rm N}\] The force $F_p$ and displacement are both parallel, so their product get the work done by $F_p$ \[W_p=F_p d=230\times 5=1150\quad {\rm J}\] (b) In the vertical path, two forces act on the box. One is the external lifting force, and the other is the force of gravity. Since the box is moving at constant speed vertically, there is no acceleration, and thus this lifting external force $F_p$ must be balanced with the weight of the box. \[F_p=F_g=mg=(1200)(10)=12000\,{\rm J}\] Assume the box is moved vertically upward. In this case, the lifting force and displacement are parallel, so the angle between them is zero $\theta=0$, and the work done by this force is \[W_p=F_p d\cos 0=12000\times 5=60\,{\rm kJ}\] On the other side, the weight force, or force of gravity $F_g=mg$ is always downward, so the angle between the box's weight and upward displacement is $180^\circ$. So, the work done by the weight of the box is \begin{align*}W_g&=F_g d\cos 180^\circ \\\\ &=(1200)(10)(5)(-1) \\\\ &=-60\,{\rm kJ}\end{align*} In such cases where the angle between $F$ and $d$ is $180^\circ$, they are called antiparallel.

Problem (7): A 40-kg crate is pushed using a force of 150 N at a distance of $6\,{\rm m}$ on a rough surface. The crate moves at a constant speed. Find (a) the work done by the external force on the crate. (b) The coefficient of kinetic friction between the crate and the floor?

Solution : (a) the crate is moved horizontally through a distance of $6\,{\rm m}$ by a force parallel to its displacement. So, the work done by this external force is \[W_p=F_p d \cos\theta=(150)(6)\cos 0=900\,{\rm J}\] where subscript $p$ denotes the person or any external agent.

(b) According to the definition of the kinetic friction force formula, $f_k=\mu_k F_N$, to find the coefficient of kinetic friction $\mu_k$, we must have both the friction force and normal force $F_N$.

In the question, we are told that the crate moves at a constant speed, so there is no acceleration, and thus, the net force applied to it must be zero.

When the friction force, which opposes the motion, is equal to the external force $F_p$, then this condition is satisfied. So, \[f_k=F_p=150\,{\rm N}\] On the other side, the crate is not lifted off the floor, so there is no motion vertically.

Balancing all forces applied vertically, the weight force and the normal force $F_N$, we can find the normal force $F_N$ as below \begin{gather*} F_N-F_g=0\\ F_N=F_g\\ \Rightarrow F_N=mg=40\times 10=400\quad {\rm N}\end{gather*} Therefore, the coefficient of kinetic friction is found to be \[f_k=\frac{f_k}{F_N}=\frac{150}{400}=0.375\]

Practice these questions to understand friction force Problems on the coefficient of friction

Problem (8): A 18-kg packing box is pulled at constant speed by a rope inclined at $20^\circ$. The box moves a distance of 20 m over a rough horizontal surface. Assume the coefficient of kinetic friction between the box and the surface to be $0.5$. (a) Find the tension in the rope? (b) How much work is done by the rope on the box?

Solution : The aim of this problem is to find the work done by the tension in the rope. The magnitude of the tension in the rope is not given. So, we must first find it.

(a) We are told the box moves at a constant speed, so, as previously mentioned, the net force on the box must be zero to produce no acceleration. But what forces are acting horizontally on the box? The horizontal component of tension in the rope, $T_{\parallel}=T\cos\theta$, and the kinetic friction force $f_k$ in the opposite direction of motion are the forces acting on the box horizontally.

If these two forces are equal in magnitude but opposite in direction, then their resultant (net) becomes zero, and consequently, the box will move at a constant speed. \begin{align*} f_k&=T_{\parallel}\\\mu_k F_N&=T\cos\theta\quad (I) \end{align*} The forces in the vertical direction must also cancel each other since there is no motion vertically. As you can see in the figure, we have \[F_N=T\sin\theta+F_g\] Substituting this into the relation (I), rearranging and solving for $T$, yields \begin{gather*} \mu_k (T\sin\theta+mg)=T\cos\theta \\\\ \Rightarrow T=\frac{\mu_k mg}{\cos\theta-\mu_k\sin\theta}\end{gather*} Substituting the numerical values into the above expression, we find the tension in the rope. \[T=\frac{(0.5)(18)(10)}{\cos 20^\circ-(0.5) \sin20^\circ}=117\quad {\rm N}\] (b) The only force that causes the box to move some distance is the horizontal component of the tension in the rope, $T_{\parallel}=T\cos\theta$. So, the work done by the tension in the rope is \begin{align*} W&=T_{\parallel}d\\ &=(117)( \cos 20^\circ)(20) \\&=2199\quad {\rm J}\end{align*}

Problem (9): A table of mass 40 kg is accelerated from rest at a constant rate of $2\,{\rm m/s^2}$ for $4\,{\rm s}$ by a constant force. What is the net work done on the table?

Solution : This is a combination of a kinematics problem and a physics work problem. Here, first, we must find the distance over which the box is displaced. The given information is: initial speed $v_0=0$, acceleration $2\,{\rm m/s^2}$, time taken $t=4\,{\rm s}$. Using this data, we can find the total displacement by applying the kinematics equation $\Delta x=\frac 12 at^2+v_0t$, \begin{align*} \Delta x&=\frac 12 at^2+v_0t\\\\&=\frac 12 (2)(4)^2+0(4)\\\\&=16\quad {\rm m}\end{align*} So, this constant force causes the table to move a distance of 16 meters across the surface. To find the work done, we need a force, as well. The force is mass times acceleration, $F=ma$, so we have \[F=ma=40\times 2=80\,{\rm N}\] Now that we have both the force and displacement, the net work done on the table is the product of force along the displacement times the magnitude of displacement \[W=80\times 16=128\quad {\rm J}\]

Work problems in a uniform circular motion

Problem (10): A 5-kg object is held at the end of a string and undergoes uniform circular motion around a circle of radius $5\,{\rm m}$. If the tangential speed of the object around the circle is $15\,{\rm m/s}$, how much work was done on the object by the centripetal force?

Solution : Here, an object moves around a circle, so we encounter a uniform circular motion problem .

In such motions around a curve or circle, that force in the radial direction exerting on the object is called the centripetal force.

On the other hand, a movement around a circle is tangent to the path at any instant of time. Thus, we conclude that in any uniform circular motion, a force is applied to the whirling object that is perpendicular to its motion at any moment of time.

So, the angle between the centripetal force and displacement at any instant is always zero, $\theta=0$. Using the work formula $W=Fd\cos\theta$, we find that the work done by the centripetal force is always zero.

This is another example of zero work in physics.

Work problems on an incline

Problem (11): A $5-{\rm kg}$ box, initially at rest, slides $2.5\,{\rm m}$ down a ramp of angle $30^\circ$. The coefficient of friction between the box and the incline is $\mu_k=0.435$. Determine (a) the work done by the gravity force, (b) the work done by the frictional force, and (c) the work done by the normal force exerted by the surface.

Solution : This part is related to problems on inclined plane surfaces . The forces acting on a box on an inclined plane are shown in the figure. As you can see, the forces along the direction of motion are the parallel component of the weight $W_{\parallel}$, and the friction force $f_k$.

(a) In the figure, you realize that the angle between the object's weight (the same as the force of gravity) and downward displacement $d$ is zero, $\theta=30^\circ$.

So, the work done by the force of gravity on the box is found using work formula as below \begin{align*} W&=Fd\cos\theta\\&=(mg)(d) \cos 30^\circ\\&=(5\times 10)(2.5) \cos 30^\circ\\&=108.25\quad {\rm J}\end{align*} (b) To find the work done by friction, we need to know its magnitude. From the kinetic friction force formula, $f_k=\mu_k F_N$, we must determine, first, the normal force acting on the box.

There is no motion in the direction perpendicular to the incline, so the resultant of forces acting in this direction must be zero. Equating the same direction forces, we will have \[F_N=mg\sin\alpha=(5)(10) \sin 30^\circ=25\,{\rm N}\] Substituting this into the above equation for kinetic friction, we can find its magnitude as \[f_k=\mu_k F_N=(0.435)(25)=10.875\,{\rm N}\] The friction force and the displacement of the box down the ramp are parallel, i.e., $\theta=0$. The work done by friction is \begin{align*} W_f&=f_kd\cos\theta \\\\ &=(10.875)(2.5) \cos 0\\\\ &=27.1875\,{\rm J}\end{align*} (c) By definition, the normal force is the same contact force that is applied to the object from the surface perpendicularly. On the other hand, the object moves along the incline, so its displacement is perpendicular to the normal force, $\theta=90^\circ$. Hence, the work done by the normal force is zero. \[W_N=F_N d\cos\theta=F_N d\cos 90^\circ=0\]

Problem (12): We want to push a $950-{\rm kg}$ heavy object 650 m up along a $7^\circ$ incline at a constant speed. How much work do we do over this distance? Ignore friction.

Solution : When it comes to constant speed in all work problems in physics, you must remember that all forces in the same direction must be equal to the opposing forces. This condition ensures that there is no acceleration in the motion.

In this case, all forces acting on the object are: the pushing force along the incline upward $F_p$, and the parallel component of the force of gravity (weight) along the incline downward, $W_{\parallel}=mg\sin\alpha$. Thus, we can find the pushing force as \begin{align*}F_p&=mg\sin 7^\circ\\\\ &=(950)(10) \sin 7^\circ \\\\& =1159\quad {\rm N}\end{align*} In the question, we are told that the object is moving up the incline, so the angle between its displacement and upward pushing force is zero, $\theta=0$. Hence, the work done by the person to push the object along the incline upward is \begin{align*} W_p&=F_p d\cos\theta\\&=1159\times 650 \cos 0\\&=753350\quad{\rm J}\end{align*}

Problem (13): Consider an electron moving at a constant speed of $1.1\times 10^6\,\rm m/s$ in a straight line. How much energy is required to stop this electron? (Take the electron's mass, $m_e=9.11\times 10^{-31}\,\rm kg$.

Solution : In this problem on work, we cannot use the work formula directly, since none of the work variables, i.e., $F$, $d$, $\theta$, are given except the velocity. In these cases, we have a problem on the work-energy theorem .

According to this rule, the net work done over a distance by a constant force on an object of mass $m$ equals the change in its kinetic energy \[W_{net}=\underbrace{\frac 12 mv_f^2-\frac 12 mv_i^2}_{\Delta K}\] Substituting the numerical values given in this problem, we get the required work to stop this fast-moving electron. \begin{align*}W_{net}&=\frac 12 m(v_f^2-v_i^2) \\\\ &=\frac 12 (9.11\times 10^{-31}) \left(0^2-(1.1\times 10^6)^2 \right) \\\\ &=-5.5\times 10^{-19}\,\rm J\end{align*} where we set the final velocity $v_f=0$ since the electron is to stop.

Problem (14): How much power is needed to lift a $25-\rm kg$ weight $1\,\rm m$ in $1\,\rm s$?

Solution : The power in physics is defined as the ratio of work done on an object to the time taken $P=\frac{W}{t}$. The SI unit of power is the watt ($W$).

In this problem, first, we must find the amount of work done in lifting the object as much as $1,\rm m$ vertically. The only force involved in this situation is the downward weight force. Thus, \[W=(mg)h=(25\times 10)(1)=250\,{\rm J}\] This amount of work has been done in a time interval of $1\,\rm s$. Hence, the power is calculated as below \[P=\frac{W}{t}=\frac{250}{1}=250\,\rm W\]

Problem (15): A particle having charge $-3.6\,\rm nC$ is released from rest in a uniform electric field $E$ moves a distance of $5\,\rm cm$ through it. The electric potential difference between those two points is $\Delta V=+400\,\rm V$. What work was done by the electric force on the particle?

Solution : The work done by the electric force on a charged particle is calculated by $W_E=qEd$, where $E$ is the magnitude of the electric field and $d$ is the amount of distance traveled through $E$. But in this case, the electric field strength is not given, and we cannot use this formula.

We can see this as a problem on electric potential . Recall that the work done by the electric force on a charge to move it between two points with different potentials is given by $W=-q\Delta V$. Substituting the given numerical values into this, we will have \begin{align*} W&=-q\Delta V \\&=-(-3.6)(+400) \\&=\boxed{1440\,\rm J} \end{align*}

Here, we learned how to calculate the work done by a constant force in physics by solving a couple of example problems.

Overall, the work done by a constant force is the product of the horizontal component of the force times the displacement between the initial and final points.

In addition, power, a related quantity to work in physics, is also defined as the rate at which work is done.

Author : Dr. Ali Nemati Date Published : 9/20/2021

TPC and eLearning
What's NEW at TPC?
Read Watch Interact
Practice Review Test
Teacher-Tools
Request a Demo
Get A Quote
Subscription Selection
Seat Calculator
Ad Free Account
Edit Profile Settings
Metric Conversions Questions
Metric System Questions
Metric Estimation Questions
Significant Digits Questions
Proportional Reasoning
Acceleration
Distance-Displacement
Dots and Graphs
Graph That Motion
Match That Graph
Name That Motion
Motion Diagrams
Pos'n Time Graphs Numerical
Pos'n Time Graphs Conceptual
Up And Down - Questions
Balanced vs. Unbalanced Forces
Change of State
Force and Motion
Mass and Weight
Match That Free-Body Diagram
Net Force (and Acceleration) Ranking Tasks
Newton's Second Law
Normal Force Card Sort
Recognizing Forces
Air Resistance and Skydiving
Solve It! with Newton's Second Law
Which One Doesn't Belong?
Component Addition Questions
Head-to-Tail Vector Addition
Projectile Mathematics
Trajectory - Angle Launched Projectiles
Trajectory - Horizontally Launched Projectiles
Vector Addition
Vector Direction
Which One Doesn't Belong? Projectile Motion
Forces in 2-Dimensions
Being Impulsive About Momentum
Explosions - Law Breakers
Hit and Stick Collisions - Law Breakers
Case Studies: Impulse and Force
Impulse-Momentum Change Table
Keeping Track of Momentum - Hit and Stick
Keeping Track of Momentum - Hit and Bounce
What's Up (and Down) with KE and PE?
Energy Conservation Questions
Energy Dissipation Questions
Energy Ranking Tasks
LOL Charts (a.k.a., Energy Bar Charts)
Match That Bar Chart
Words and Charts Questions
Name That Energy
Stepping Up with PE and KE Questions
Case Studies - Circular Motion
Circular Logic
Forces and Free-Body Diagrams in Circular Motion
Gravitational Field Strength
Universal Gravitation
Angular Position and Displacement
Linear and Angular Velocity
Angular Acceleration
Rotational Inertia
Balanced vs. Unbalanced Torques
Getting a Handle on Torque
Torque-ing About Rotation
Properties of Matter
Fluid Pressure
Buoyant Force
Sinking, Floating, and Hanging
Pascal's Principle
Flow Velocity
Bernoulli's Principle
Balloon Interactions
Charge and Charging
Charge Interactions
Charging by Induction
Conductors and Insulators
Coulombs Law
Electric Field
Electric Field Intensity
Polarization
Case Studies: Electric Power
Know Your Potential
Light Bulb Anatomy
I = ∆V/R Equations as a Guide to Thinking
Parallel Circuits - ∆V = I•R Calculations
Resistance Ranking Tasks
Series Circuits - ∆V = I•R Calculations
Series vs. Parallel Circuits
Equivalent Resistance
Period and Frequency of a Pendulum
Pendulum Motion: Velocity and Force
Energy of a Pendulum
Period and Frequency of a Mass on a Spring
Horizontal Springs: Velocity and Force
Vertical Springs: Velocity and Force
Energy of a Mass on a Spring
Decibel Scale
Frequency and Period
Closed-End Air Columns
Name That Harmonic: Strings
Rocking the Boat
Wave Basics
Matching Pairs: Wave Characteristics
Wave Interference
Waves - Case Studies
Color Addition and Subtraction
Color Filters
If This, Then That: Color Subtraction
Light Intensity
Color Pigments
Converging Lenses
Curved Mirror Images
Law of Reflection
Refraction and Lenses
Total Internal Reflection
Who Can See Who?
Lab Equipment
Lab Procedures
Formulas and Atom Counting
Atomic Models
Bond Polarity
Entropy Questions
Cell Voltage Questions
Heat of Formation Questions
Reduction Potential Questions
Oxidation States Questions
Measuring the Quantity of Heat
Hess's Law
Oxidation-Reduction Questions
Galvanic Cells Questions
Thermal Stoichiometry
Molecular Polarity
Quantum Mechanics
Balancing Chemical Equations
Bronsted-Lowry Model of Acids and Bases
Classification of Matter
Collision Model of Reaction Rates
Density Ranking Tasks
Dissociation Reactions
Complete Electron Configurations
Elemental Measures
Enthalpy Change Questions
Equilibrium Concept
Equilibrium Constant Expression
Equilibrium Calculations - Questions
Equilibrium ICE Table
Intermolecular Forces Questions
Ionic Bonding
Lewis Electron Dot Structures
Limiting Reactants
Line Spectra Questions
Mass Stoichiometry
Measurement and Numbers
Metals, Nonmetals, and Metalloids
Metric Estimations
Metric System
Molarity Ranking Tasks
Mole Conversions
Name That Element
Names to Formulas
Names to Formulas 2
Nuclear Decay
Particles, Words, and Formulas
Periodic Trends
Precipitation Reactions and Net Ionic Equations
Pressure Concepts
Pressure-Temperature Gas Law
Pressure-Volume Gas Law
Chemical Reaction Types
Significant Digits and Measurement
States Of Matter Exercise
Stoichiometry Law Breakers
Stoichiometry - Math Relationships
Subatomic Particles
Spontaneity and Driving Forces
Gibbs Free Energy
Volume-Temperature Gas Law
Acid-Base Properties
Energy and Chemical Reactions
Chemical and Physical Properties
Valence Shell Electron Pair Repulsion Theory
Writing Balanced Chemical Equations
Mission CG1
Mission CG10
Mission CG2
Mission CG3
Mission CG4
Mission CG5
Mission CG6
Mission CG7
Mission CG8
Mission CG9
Mission EC1
Mission EC10
Mission EC11
Mission EC12
Mission EC2
Mission EC3
Mission EC4
Mission EC5
Mission EC6
Mission EC7
Mission EC8
Mission EC9
Mission RL1
Mission RL2
Mission RL3
Mission RL4
Mission RL5
Mission RL6
Mission KG7
Mission RL8
Mission KG9
Mission RL10
Mission RL11
Mission RM1
Mission RM2
Mission RM3
Mission RM4
Mission RM5
Mission RM6
Mission RM8
Mission RM10
Mission LC1
Mission RM11
Mission LC2
Mission LC3
Mission LC4
Mission LC5
Mission LC6
Mission LC8
Mission SM1
Mission SM2
Mission SM3
Mission SM4
Mission SM5
Mission SM6
Mission SM8
Mission SM10
Mission KG10
Mission SM11
Mission KG2
Mission KG3
Mission KG4
Mission KG5
Mission KG6
Mission KG8
Mission KG11
Mission F2D1
Mission F2D2
Mission F2D3
Mission F2D4
Mission F2D5
Mission F2D6
Mission KC1
Mission KC2
Mission KC3
Mission KC4
Mission KC5
Mission KC6
Mission KC7
Mission KC8
Mission AAA
Mission SM9
Mission LC7
Mission LC9
Mission NL1
Mission NL2
Mission NL3
Mission NL4
Mission NL5
Mission NL6
Mission NL7
Mission NL8
Mission NL9
Mission NL10
Mission NL11
Mission NL12
Mission MC1
Mission MC10
Mission MC2
Mission MC3
Mission MC4
Mission MC5
Mission MC6
Mission MC7
Mission MC8
Mission MC9
Mission RM7
Mission RM9
Mission RL7
Mission RL9
Mission SM7
Mission SE1
Mission SE10
Mission SE11
Mission SE12
Mission SE2
Mission SE3
Mission SE4
Mission SE5
Mission SE6
Mission SE7
Mission SE8
Mission SE9
Mission VP1
Mission VP10
Mission VP2
Mission VP3
Mission VP4
Mission VP5
Mission VP6
Mission VP7
Mission VP8
Mission VP9
Mission WM1
Mission WM2
Mission WM3
Mission WM4
Mission WM5
Mission WM6
Mission WM7
Mission WM8
Mission WE1
Mission WE10
Mission WE2
Mission WE3
Mission WE4
Mission WE5
Mission WE6
Mission WE7
Mission WE8
Mission WE9
Vector Walk Interactive
Name That Motion Interactive
Kinematic Graphing 1 Concept Checker
Kinematic Graphing 2 Concept Checker
Graph That Motion Interactive
Two Stage Rocket Interactive
Rocket Sled Concept Checker
Force Concept Checker
Free-Body Diagrams Concept Checker
Free-Body Diagrams The Sequel Concept Checker
Skydiving Concept Checker
Elevator Ride Concept Checker
Vector Addition Concept Checker
Vector Walk in Two Dimensions Interactive
Name That Vector Interactive
River Boat Simulator Concept Checker
Projectile Simulator 2 Concept Checker
Projectile Simulator 3 Concept Checker
Hit the Target Interactive
Turd the Target 1 Interactive
Turd the Target 2 Interactive
Balance It Interactive
Go For The Gold Interactive
Egg Drop Concept Checker
Fish Catch Concept Checker
Exploding Carts Concept Checker
Collision Carts - Inelastic Collisions Concept Checker
Its All Uphill Concept Checker
Stopping Distance Concept Checker
Chart That Motion Interactive
Roller Coaster Model Concept Checker
Uniform Circular Motion Concept Checker
Horizontal Circle Simulation Concept Checker
Vertical Circle Simulation Concept Checker
Race Track Concept Checker
Gravitational Fields Concept Checker
Orbital Motion Concept Checker
Angular Acceleration Concept Checker
Balance Beam Concept Checker
Torque Balancer Concept Checker
Aluminum Can Polarization Concept Checker
Charging Concept Checker
Name That Charge Simulation
Coulomb's Law Concept Checker
Electric Field Lines Concept Checker
Put the Charge in the Goal Concept Checker
Circuit Builder Concept Checker (Series Circuits)
Circuit Builder Concept Checker (Parallel Circuits)
Circuit Builder Concept Checker (∆V-I-R)
Circuit Builder Concept Checker (Voltage Drop)
Equivalent Resistance Interactive
Pendulum Motion Simulation Concept Checker
Mass on a Spring Simulation Concept Checker
Particle Wave Simulation Concept Checker
Boundary Behavior Simulation Concept Checker
Slinky Wave Simulator Concept Checker
Simple Wave Simulator Concept Checker
Wave Addition Simulation Concept Checker
Standing Wave Maker Simulation Concept Checker
Color Addition Concept Checker
Painting With CMY Concept Checker
Stage Lighting Concept Checker
Filtering Away Concept Checker
InterferencePatterns Concept Checker
Young's Experiment Interactive
Plane Mirror Images Interactive
Who Can See Who Concept Checker
Optics Bench (Mirrors) Concept Checker
Name That Image (Mirrors) Interactive
Refraction Concept Checker
Total Internal Reflection Concept Checker
Optics Bench (Lenses) Concept Checker
Kinematics Preview
Velocity Time Graphs Preview
Moving Cart on an Inclined Plane Preview
Stopping Distance Preview
Cart, Bricks, and Bands Preview
Fan Cart Study Preview
Friction Preview
Coffee Filter Lab Preview
Friction, Speed, and Stopping Distance Preview
Up and Down Preview
Projectile Range Preview
Ballistics Preview
Juggling Preview
Marshmallow Launcher Preview
Air Bag Safety Preview
Colliding Carts Preview
Collisions Preview
Engineering Safer Helmets Preview
Push the Plow Preview
Its All Uphill Preview
Energy on an Incline Preview
Modeling Roller Coasters Preview
Hot Wheels Stopping Distance Preview
Ball Bat Collision Preview
Energy in Fields Preview
Weightlessness Training Preview
Roller Coaster Loops Preview
Universal Gravitation Preview
Keplers Laws Preview
Kepler's Third Law Preview
Charge Interactions Preview
Sticky Tape Experiments Preview
Wire Gauge Preview
Voltage, Current, and Resistance Preview
Light Bulb Resistance Preview
Series and Parallel Circuits Preview
Thermal Equilibrium Preview
Linear Expansion Preview
Heating Curves Preview
Electricity and Magnetism - Part 1 Preview
Electricity and Magnetism - Part 2 Preview
Vibrating Mass on a Spring Preview
Period of a Pendulum Preview
Wave Speed Preview
Slinky-Experiments Preview
Standing Waves in a Rope Preview
Sound as a Pressure Wave Preview
DeciBel Scale Preview
DeciBels, Phons, and Sones Preview
Sound of Music Preview
Shedding Light on Light Bulbs Preview
Models of Light Preview
Electromagnetic Radiation Preview
Electromagnetic Spectrum Preview
EM Wave Communication Preview
Digitized Data Preview
Light Intensity Preview
Concave Mirrors Preview
Object Image Relations Preview
Snells Law Preview
Reflection vs. Transmission Preview
Magnification Lab Preview
Reactivity Preview
Ions and the Periodic Table Preview
Periodic Trends Preview
Chemical Reactions Preview
Intermolecular Forces Preview
Melting Points and Boiling Points Preview
Bond Energy and Reactions Preview
Reaction Rates Preview
Ammonia Factory Preview
Stoichiometry Preview
Nuclear Chemistry Preview
Gaining Teacher Access
Task Tracker Directions
Conceptual Physics Course
On-Level Physics Course
Honors Physics Course
Chemistry Concept Builders
All Chemistry Resources
Users Voice
Tasks and Classes
Webinars and Trainings
Subscription
Subscription Locator
1-D Kinematics
Newton's Laws
Vectors - Motion and Forces in Two Dimensions
Momentum and Its Conservation
Work and Energy
Circular Motion and Satellite Motion
Thermal Physics
Static Electricity
Electric Circuits
Vibrations and Waves
Sound Waves and Music
Light and Color
Reflection and Mirrors
Measurement and Calculations
About the Physics Interactives
Task Tracker
Usage Policy
Newtons Laws
Vectors and Projectiles
Forces in 2D
Momentum and Collisions
Circular and Satellite Motion
Balance and Rotation
Electromagnetism
Waves and Sound
Atomic Physics
Forces in Two Dimensions
Work, Energy, and Power
Circular Motion and Gravitation
Sound Waves
1-Dimensional Kinematics
Circular, Satellite, and Rotational Motion
Einstein's Theory of Special Relativity
Waves, Sound and Light
QuickTime Movies
About the Concept Builders
Pricing For Schools
Directions for Version 2
Measurement and Units
Relationships and Graphs
Rotation and Balance
Vibrational Motion
Reflection and Refraction
Teacher Accounts
Kinematic Concepts
Kinematic Graphing
Wave Motion
Sound and Music
About CalcPad
1D Kinematics
Vectors and Forces in 2D
Simple Harmonic Motion
Rotational Kinematics
Rotation and Torque
Rotational Dynamics
Electric Fields, Potential, and Capacitance
Transient RC Circuits
Light Waves
Units and Measurement
Stoichiometry
Molarity and Solutions
Thermal Chemistry
Acids and Bases
Kinetics and Equilibrium
Solution Equilibria
Oxidation-Reduction
Nuclear Chemistry
Newton's Laws of Motion
Work and Energy Packet
Static Electricity Review
NGSS Alignments
1D-Kinematics
Projectiles
Circular Motion
Magnetism and Electromagnetism
Graphing Practice
About the ACT
ACT Preparation
For Teachers
Other Resources
Solutions Guide
Solutions Guide Digital Download
Motion in One Dimension
Work, Energy and Power
Chemistry of Matter
Names and Formulas
Algebra Based On-Level Physics
Honors Physics
Conceptual Physics
Other Tools
Frequently Asked Questions
Purchasing the Download
Purchasing the Digital Download
About the NGSS Corner
NGSS Search
Force and Motion DCIs - High School
Energy DCIs - High School
Wave Applications DCIs - High School
Force and Motion PEs - High School
Energy PEs - High School
Wave Applications PEs - High School
Crosscutting Concepts
The Practices
Physics Topics
NGSS Corner: Activity List
NGSS Corner: Infographics
About the Toolkits
Position-Velocity-Acceleration
Position-Time Graphs
Velocity-Time Graphs
Newton's First Law
Newton's Second Law
Newton's Third Law
Terminal Velocity
Projectile Motion
Forces in 2 Dimensions
Impulse and Momentum Change
Momentum Conservation
Work-Energy Fundamentals
Work-Energy Relationship
Roller Coaster Physics
Satellite Motion
Electric Fields
Circuit Concepts
Series Circuits
Parallel Circuits
Describing-Waves
Wave Behavior Toolkit
Standing Wave Patterns
Resonating Air Columns
Wave Model of Light
Plane Mirrors
Curved Mirrors
Teacher Guide
Using Lab Notebooks
Current Electricity
Light Waves and Color
Reflection and Ray Model of Light
Refraction and Ray Model of Light
Teacher Resources
Subscriptions

Newton's Laws
Einstein's Theory of Special Relativity
About Concept Checkers
School Pricing
Newton's Laws of Motion
Newton's First Law
Newton's Third Law

Mechanics: Work, Energy and Power

Calculator pad, version 2, work, energy and power: problem set.

Renatta Gass is out with her friends. Misfortune occurs and Renatta and her friends find themselves getting a work out. They apply a cumulative force of 1080 N to push the car 218 m to the nearest fuel station. Determine the work done on the car.

Audio Guided Solution
Show Answer

Hans Full is pulling on a rope to drag his backpack to school across the ice. He pulls upwards and rightwards with a force of 22.9 Newtons at an angle of 35 degrees above the horizontal to drag his backpack a horizontal distance of 129 meters to the right. Determine the work (in Joules) done upon the backpack.

Lamar Gant, U.S. powerlifting star, became the first man to deadlift five times his own body weight in 1985. Deadlifting involves raising a loaded barbell from the floor to a position above the head with outstretched arms. Determine the work done by Lamar in deadlifting 300 kg to a height of 0.90 m above the ground.

Sheila has just arrived at the airport and is dragging her suitcase to the luggage check-in desk. She pulls on the strap with a force of 190 N at an angle of 35° to the horizontal to displace it 45 m to the desk. Determine the work done by Sheila on the suitcase.

While training for breeding season, a 380 gram male squirrel does 32 pushups in a minute, displacing its center of mass by a distance of 8.5 cm for each pushup. Determine the total work done on the squirrel while moving upward (32 times).

During the Powerhouse lab, Jerome runs up the stairs, elevating his 102 kg body a vertical distance of 2.29 meters in a time of 1.32 seconds at a constant speed.

a. Determine the work done by Jerome in climbing the stair case. b. Determine the power generated by Jerome.

A new conveyor system at the local packaging plan will utilize a motor-powered mechanical arm to exert an average force of 890 N to push large crates a distance of 12 meters in 22 seconds. Determine the power output required of such a motor.

The Taipei 101 in Taiwan is a 1667-foot tall, 101-story skyscraper. The skyscraper is the home of the world’s fastest elevator. The elevators transport visitors from the ground floor to the Observation Deck on the 89th floor at speeds up to 16.8 m/s. Determine the power delivered by the motor to lift the 10 passengers at this speed. The combined mass of the passengers and cabin is 1250 kg.

The ski slopes at Bluebird Mountain make use of tow ropes to transport snowboarders and skiers to the summit of the hill. One of the tow ropes is powered by a 22-kW motor which pulls skiers along an icy incline of 14° at a constant speed. Suppose that 18 skiers with an average mass of 48 kg hold onto the rope and suppose that the motor operates at full power.

a. Determine the cumulative weight of all these skiers. b. Determine the force required to pull this amount of weight up a 14° incline at a constant speed. c. Determine the speed at which the skiers will ascend the hill.

Problem 10:

The first asteroid to be discovered is Ceres. It is the largest and most massive asteroid is our solar system’s asteroid belt, having an estimated mass of 3.0 x 10 21 kg and an orbital speed of 17900 m/s. Determine the amount of kinetic energy possessed by Ceres.

Problem 11:

A bicycle has a kinetic energy of 124 J. What kinetic energy would the bicycle have if it had …

a. … twice the mass and was moving at the same speed? b. … the same mass and was moving with twice the speed? c. … one-half the mass and was moving with twice the speed? d. … the same mass and was moving with one-half the speed? e. … three times the mass and was moving with one-half the speed?

Problem 12:

A 78-kg skydiver has a speed of 62 m/s at an altitude of 870 m above the ground.

a. Determine the kinetic energy possessed by the skydiver. b. Determine the potential energy possessed by the skydiver. c. Determine the total mechanical energy possessed by the skydiver.

Problem 13:

Lee Ben Fardest (esteemed American ski jumper), has a mass of 59.6 kg. He is moving with a speed of 23.4 m/s at a height of 44.6 meters above the ground. Determine the total mechanical energy of Lee Ben Fardest.

Problem 14:

Chloe leads South’s varsity softball team in hitting. In a game against New Greer Academy this past weekend, Chloe slugged the 181-gram softball so hard that it cleared the outfield fence and landed on Lake Avenue. At one point in its trajectory, the ball was 28.8 m above the ground and moving with a speed of 19.7 m/s. Determine the total mechanical energy of the softball.

Problem 15:

Olive Udadi is at the park with her father. The 26-kg Olive is on a swing following the path as shown. Olive has a speed of 0 m/s at position A and is a height of 3.0-m above the ground. At position B, Olive is 1.2 m above the ground. At position C (2.2 m above the ground), Olive projects from the seat and travels as a projectile along the path shown. At point F, Olive is a mere picometer above the ground. Assume negligible air resistance throughout the motion. Use this information to fill in the table.


	3.0				0.0
	1.2
	2.2
	0


	3.0				0.0
	1.2
	2.2
	0

Problem 16:

Suzie Lavtaski (m=56 kg) is skiing at Bluebird Mountain. She is moving at 16 m/s across the crest of a ski hill located 34 m above ground level at the end of the run.

a. Determine Suzie's kinetic energy. b. Determine Suzie's potential energy relative to the height of the ground at the end of the run. c. Determine Suzie's total mechanical energy at the crest of the hill. d. If no energy is lost or gained between the top of the hill and her initial arrival at the end of the run, then what will be Suzie's total mechanical energy at the end of the run? e. Determine Suzie's speed as she arrives at the end of the run and prior to braking to a stop.

Problem 17:

Nicholas is at The Noah's Ark Amusement Park and preparing to ride on The Point of No Return racing slide. At the top of the slide, Nicholas (m=72.6 kg) is 28.5 m above the ground.

a. Determine Nicholas' potential energy at the top of the slide. b. Determine Nicholas's kinetic energy at the top of the slide. c. Assuming negligible losses of energy between the top of the slide and his approach to the bottom of the slide (h=0 m), determine Nicholas's total mechanical energy as he arrives at the bottom of the slide. d. Determine Nicholas' potential energy as he arrives at the bottom of the slide. e. Determine Nicholas' kinetic energy as he arrives at the bottom of the slide. f. Determine Nicholas' speed as he arrives at the bottom of the slide.

Problem 18:

Ima Scaarred (m=56.2 kg) is traveling at a speed of 12.8 m/s at the top of a 19.5-m high roller coaster loop.

a. Determine Ima's kinetic energy at the top of the loop. b. Determine Ima's potential energy at the top of the loop. c. Assuming negligible losses of energy due to friction and air resistance, determine Ima's total mechanical energy at the bottom of the loop (h=0 m). d. Determine Ima's speed at the bottom of the loop.

Problem 19:

Justin Thyme is traveling down Lake Avenue at 32.8 m/s in his 1510-kg 1992 Camaro. He spots a police car with a radar gun and quickly slows down to a legal speed of 20.1 m/s.

a. Determine the initial kinetic energy of the Camaro. b. Determine the kinetic energy of the Camaro after slowing down. c. Determine the amount of work done on the Camaro during the deceleration.

Problem 20:

Pete Zaria works on weekends at Barnaby's Pizza Parlor. His primary responsibility is to fill drink orders for customers. He fills a pitcher full of Cola, places it on the counter top and gives the 2.6-kg pitcher a 8.8 N forward push over a distance of 48 cm to send it to a customer at the end of the counter. The coefficient of friction between the pitcher and the counter top is 0.28.

a. Determine the work done by Pete on the pitcher during the 48 cm push. b. Determine the work done by friction upon the pitcher . c. Determine the total work done upon the pitcher . d. Determine the kinetic energy of the pitcher when Pete is done pushing it. e. Determine the speed of the pitcher when Pete is done pushing it.

Problem 21:

The Top Thrill Dragster stratacoaster at Cedar Point Amusement Park in Ohio uses a hydraulic launching system to accelerate riders from 0 to 53.6 m/s (120 mi/hr) in 3.8 seconds before climbing a completely vertical 420-foot hill.

a. Jerome (m=102 kg) visits the park with his church youth group. He boards his car, straps himself in and prepares for the thrill of the day. What is Jerome's kinetic energy before the acceleration period? b. The 3.8-second acceleration period begins to accelerate Jerome along the level track. What is Jerome's kinetic energy at the end of this acceleration period? c. Once the launch is over, Jerome begins screaming up the 420-foot, completely vertical section of the track. Determine Jerome's potential energy at the top of the vertical section. ( GIVEN : 1.00 m = 3.28 ft) d. Determine Jerome's kinetic energy at the top of the vertical section. e. Determine Jerome's speed at the top of the vertical section.

Problem 22:

Paige is the tallest player on South's Varsity volleyball team. She is in spiking position when Julia gives her the perfect set. The 0.226-kg volleyball is 2.29 m above the ground and has a speed of 1.06 m/s. Paige spikes the ball, doing 9.89 J of work on it.

a. Determine the potential energy of the ball before Paige spikes it. b. Determine the kinetic energy of the ball before Paige spikes it. c. Determine the total mechanical energy of the ball before Paige spikes it. d. Determine the total mechanical energy of the ball upon hitting the floor on the opponent's side of the net. e. Determine the speed of the ball upon hitting the floor on the opponent's side of the net.

Problem 23:

According to ABC's Wide World of Sports show, there is the thrill of victory and the agony of defeat. On March 21 of 1970, Vinko Bogataj was the Yugoslavian entrant into the World Championships held in former West Germany. By his third and final jump of the day, heavy and persistent snow produced dangerous conditions along the slope. Midway through the run, Bogataj recognized the danger and attempted to make adjustments in order to terminate his jump. Instead, he lost his balanced and tumbled and flipped off the slope into the dense crowd. For nearly 30 years thereafter, footage of the event was included in the introduction of ABC's infamous sports show and Vinco has become known as the agony of defeat icon.

a. Determine the speed of 72-kg Vinco after skiing down the hill to a height which is 49 m below the starting location. b. After descending the 49 m, Vinko tumbled off the track and descended another 15 m down the ski hill before finally stopping. Determine the change in potential energy of Vinko from the top of the hill to the point at which he stops. c. Determine the amount of cumulative work done upon Vinko's body as he crashes to a halt.

Problem 24:

Nolan Ryan reportedly had the fastest pitch in baseball, clocked at 100.9 mi/hr (45.0 m/s) If such a pitch had been directed vertically upwards at this same speed, then to what height would it have traveled?

Problem 25:

In the Incline Energy lab, partners Anna Litical and Noah Formula give a 1.00-kg cart an initial speed of 2.35 m/s from a height of 0.125 m above the lab table. Determine the speed of the cart when it is located 0.340 m above the lab table.

Problem 26:

In April of 1976, Chicago Cub slugger Dave Kingman hit a home run which cleared the Wrigley Field fence and hit a house located 530 feet (162 m) from home plate. Suppose that the 0.145-kg baseball left Kingman's bat at 92.7 m/s and that it lost 10% of its original energy on its flight through the air. Determine the speed of the ball when it cleared the stadium wall at a height of 25.6 m.

Problem 27:

Dizzy is speeding along at 22.8 m/s as she approaches the level section of track near the loading dock of the Whizzer roller coaster ride. A braking system abruptly brings the 328-kg car (rider mass included) to a speed of 2.9 m/s over a distance of 5.55 meters. Determine the braking force applied to Dizzy's car.

Problem 28:

A 6.8-kg toboggan is kicked on a frozen pond, such that it acquires a speed of 1.9 m/s. The coefficient of friction between the pond and the toboggan is 0.13. Determine the distance which the toboggan slides before coming to rest.

Problem 29:

Connor (m=76.0 kg) is competing in the state diving championship. He leaves the springboard from a height of 3.00 m above the water surface with a speed of 5.94 m/s in the upward direction. a. Determine Connor's speed when he strikes the water. b. Connor's body plunges to a depth of 2.15 m below the water surface before stopping. Determine the average force of water resistance experienced by his body.

Problem 30:

Gwen is baby-sitting for the Parker family. She takes 3-year old Allison to the neighborhood park and places her in the seat of the children's swing. Gwen pulls the 1.8-m long chain back to make a 26° angle with the vertical and lets 14-kg Allison (swing mass included) go. Assuming negligible friction and air resistance, determine Allison's speed at the lowest point in the trajectory.

Problem 31:

Sheila (m=56.8 kg) is in her saucer sled moving at 12.6 m/s at the bottom of the sledding hill near Bluebird Lake. She approaches a long embankment inclined upward at 16° above the horizontal. As she slides up the embankment, she encounters a coefficient of friction of 0.128. Determine the height to which she will travel before coming to rest.

Problem 32:

Matthew starts from rest on top of 8.45 m high sledding hill. He slides down the 32-degree incline and across the plateau at its base. The coefficient of friction between the sled and snow is 0.128 for both the hill and the plateau. Matthew and the sled have a combined mass of 27.5 kg. Determine the distance which Matthew will slide along the level surface before coming to a complete stop.

Return to Overview

View Audio Guided Solution for Problem:

40 problem-solving techniques and processes

All teams and organizations encounter challenges. Approaching those challenges without a structured problem solving process can end up making things worse.

Proven problem solving techniques such as those outlined below can guide your group through a process of identifying problems and challenges , ideating on possible solutions , and then evaluating and implementing the most suitable .

In this post, you'll find problem-solving tools you can use to develop effective solutions. You'll also find some tips for facilitating the problem solving process and solving complex problems.

Design your next session with SessionLab

Join the 150,000+ facilitators  using SessionLab.

What is problem solving?

Problem solving is a process of finding and implementing a solution to a challenge or obstacle. In most contexts, this means going through a problem solving process that begins with identifying the issue, exploring its root causes, ideating and refining possible solutions before implementing and measuring the impact of that solution.

For simple or small problems, it can be tempting to skip straight to implementing what you believe is the right solution. The danger with this approach is that without exploring the true causes of the issue, it might just occur again or your chosen solution may cause other issues.

Particularly in the world of work, good problem solving means using data to back up each step of the process, bringing in new perspectives and effectively measuring the impact of your solution.

Effective problem solving can help ensure that your team or organization is well positioned to overcome challenges, be resilient to change and create innovation. In my experience, problem solving is a combination of skillset, mindset and process, and it’s especially vital for leaders to cultivate this skill.

A group of people looking at a poster with notes on it

What is the seven step problem solving process?

A problem solving process is a step-by-step framework from going from discovering a problem all the way through to implementing a solution.

With practice, this framework can become intuitive, and innovative companies tend to have a consistent and ongoing ability to discover and tackle challenges when they come up.

You might see everything from a four step problem solving process through to seven steps. While all these processes cover roughly the same ground, I’ve found a seven step problem solving process is helpful for making all key steps legible.

We’ll outline that process here and then follow with techniques you can use to explore and work on that step of the problem solving process with a group.

The seven-step problem solving process is:

1. Problem identification

The first stage of any problem solving process is to identify the problem(s) you need to solve. This often looks like using group discussions and activities to help a group surface and effectively articulate the challenges they’re facing and wish to resolve.

Be sure to align with your team on the exact definition and nature of the problem you’re solving. An effective process is one where everyone is pulling in the same direction – ensure clarity and alignment now to help avoid misunderstandings later.

2. Problem analysis and refinement

The process of problem analysis means ensuring that the problem you are seeking to solve is the right problem . Choosing the right problem to solve means you are on the right path to creating the right solution.

At this stage, you may look deeper at the problem you identified to try and discover the root cause at the level of people or process. You may also spend some time sourcing data, consulting relevant parties and creating and refining a problem statement.

Problem refinement means adjusting scope or focus of the problem you will be aiming to solve based on what comes up during your analysis. As you analyze data sources, you might discover that the root cause means you need to adjust your problem statement. Alternatively, you might find that your original problem statement is too big to be meaningful approached within your current project.

Remember that the goal of any problem refinement is to help set the stage for effective solution development and deployment. Set the right focus and get buy-in from your team here and you’ll be well positioned to move forward with confidence.

3. Solution generation

Once your group has nailed down the particulars of the problem you wish to solve, you want to encourage a free flow of ideas connecting to solving that problem. This can take the form of problem solving games that encourage creative thinking or techniquess designed to produce working prototypes of possible solutions.

The key to ensuring the success of this stage of the problem solving process is to encourage quick, creative thinking and create an open space where all ideas are considered. The best solutions can often come from unlikely places and by using problem solving techniques that celebrate invention, you might come up with solution gold.

4. Solution development

No solution is perfect right out of the gate. It’s important to discuss and develop the solutions your group has come up with over the course of following the previous problem solving steps in order to arrive at the best possible solution. Problem solving games used in this stage involve lots of critical thinking, measuring potential effort and impact, and looking at possible solutions analytically.

During this stage, you will often ask your team to iterate and improve upon your front-running solutions and develop them further. Remember that problem solving strategies always benefit from a multitude of voices and opinions, and not to let ego get involved when it comes to choosing which solutions to develop and take further.

Finding the best solution is the goal of all problem solving workshops and here is the place to ensure that your solution is well thought out, sufficiently robust and fit for purpose.

5. Decision making and planning

Nearly there! Once you’ve got a set of possible, you’ll need to make a decision on which to implement. This can be a consensus-based group decision or it might be for a leader or major stakeholder to decide. You’ll find a set of effective decision making methods below.

Once your group has reached consensus and selected a solution, there are some additional actions that also need to be decided upon. You’ll want to work on allocating ownership of the project, figure out who will do what, how the success of the solution will be measured and decide the next course of action.

Set clear accountabilities, actions, timeframes, and follow-ups for your chosen solution. Make these decisions and set clear next-steps in the problem solving workshop so that everyone is aligned and you can move forward effectively as a group.

Ensuring that you plan for the roll-out of a solution is one of the most important problem solving steps. Without adequate planning or oversight, it can prove impossible to measure success or iterate further if the problem was not solved.

6. Solution implementation

This is what we were waiting for! All problem solving processes have the end goal of implementing an effective and impactful solution that your group has confidence in.

Project management and communication skills are key here – your solution may need to adjust when out in the wild or you might discover new challenges along the way. For some solutions, you might also implement a test with a small group and monitor results before rolling it out to an entire company.

You should have a clear owner for your solution who will oversee the plans you made together and help ensure they’re put into place. This person will often coordinate the implementation team and set-up processes to measure the efficacy of your solution too.

7. Solution evaluation

So you and your team developed a great solution to a problem and have a gut feeling it’s been solved. Work done, right? Wrong. All problem solving strategies benefit from evaluation, consideration, and feedback.

You might find that the solution does not work for everyone, might create new problems, or is potentially so successful that you will want to roll it out to larger teams or as part of other initiatives.

None of that is possible without taking the time to evaluate the success of the solution you developed in your problem solving model and adjust if necessary.

Remember that the problem solving process is often iterative and it can be common to not solve complex issues on the first try. Even when this is the case, you and your team will have generated learning that will be important for future problem solving workshops or in other parts of the organization.

It’s also worth underlining how important record keeping is throughout the problem solving process. If a solution didn’t work, you need to have the data and records to see why that was the case. If you go back to the drawing board, notes from the previous workshop can help save time.

What does an effective problem solving process look like?

Every effective problem solving process begins with an agenda . In our experience, a well-structured problem solving workshop is one of the best methods for successfully guiding a group from exploring a problem to implementing a solution.

The format of a workshop ensures that you can get buy-in from your group, encourage free-thinking and solution exploration before making a decision on what to implement following the session.

This Design Sprint 2.0 template is an effective problem solving process from top agency AJ&Smart. It’s a great format for the entire problem solving process, with four-days of workshops designed to surface issues, explore solutions and even test a solution.

Check it for an example of how you might structure and run a problem solving process and feel free to copy and adjust it your needs!

For a shorter process you can run in a single afternoon, this remote problem solving agenda will guide you effectively in just a couple of hours.

Whatever the length of your workshop, by using SessionLab, it’s easy to go from an idea to a complete agenda . Start by dragging and dropping your core problem solving activities into place . Add timings, breaks and necessary materials before sharing your agenda with your colleagues.

The resulting agenda will be your guide to an effective and productive problem solving session that will also help you stay organized on the day!

Complete problem-solving methods

In this section, we’ll look at in-depth problem-solving methods that provide a complete end-to-end process for developing effective solutions. These will help guide your team from the discovery and definition of a problem through to delivering the right solution.

If you’re looking for an all-encompassing method or problem-solving model, these processes are a great place to start. They’ll ask your team to challenge preconceived ideas and adopt a mindset for solving problems more effectively.

Six Thinking Hats

Individual approaches to solving a problem can be very different based on what team or role an individual holds. It can be easy for existing biases or perspectives to find their way into the mix, or for internal politics to direct a conversation.

Six Thinking Hats is a classic method for identifying the problems that need to be solved and enables your team to consider them from different angles, whether that is by focusing on facts and data, creative solutions, or by considering why a particular solution might not work.

Like all problem-solving frameworks, Six Thinking Hats is effective at helping teams remove roadblocks from a conversation or discussion and come to terms with all the aspects necessary to solve complex problems.

The Six Thinking Hats #creative thinking #meeting facilitation #problem solving #issue resolution #idea generation #conflict resolution The Six Thinking Hats are used by individuals and groups to separate out conflicting styles of thinking. They enable and encourage a group of people to think constructively together in exploring and implementing change, rather than using argument to fight over who is right and who is wrong.

Lightning Decision Jam

Featured courtesy of Jonathan Courtney of AJ&Smart Berlin, Lightning Decision Jam is one of those strategies that should be in every facilitation toolbox. Exploring problems and finding solutions is often creative in nature, though as with any creative process, there is the potential to lose focus and get lost.

Unstructured discussions might get you there in the end, but it’s much more effective to use a method that creates a clear process and team focus.

In Lightning Decision Jam, participants are invited to begin by writing challenges, concerns, or mistakes on post-its without discussing them before then being invited by the moderator to present them to the group.

From there, the team vote on which problems to solve and are guided through steps that will allow them to reframe those problems, create solutions and then decide what to execute on.

By deciding the problems that need to be solved as a team before moving on, this group process is great for ensuring the whole team is aligned and can take ownership over the next stages.

Lightning Decision Jam (LDJ) #action #decision making #problem solving #issue analysis #innovation #design #remote-friendly It doesn’t matter where you work and what your job role is, if you work with other people together as a team, you will always encounter the same challenges: Unclear goals and miscommunication that cause busy work and overtime Unstructured meetings that leave attendants tired, confused and without clear outcomes. Frustration builds up because internal challenges to productivity are not addressed Sudden changes in priorities lead to a loss of focus and momentum Muddled compromise takes the place of clear decision- making, leaving everybody to come up with their own interpretation. In short, a lack of structure leads to a waste of time and effort, projects that drag on for too long and frustrated, burnt out teams. AJ&Smart has worked with some of the most innovative, productive companies in the world. What sets their teams apart from others is not better tools, bigger talent or more beautiful offices. The secret sauce to becoming a more productive, more creative and happier team is simple: Replace all open discussion or brainstorming with a structured process that leads to more ideas, clearer decisions and better outcomes. When a good process provides guardrails and a clear path to follow, it becomes easier to come up with ideas, make decisions and solve problems. This is why AJ&Smart created Lightning Decision Jam (LDJ). It’s a simple and short, but powerful group exercise that can be run either in-person, in the same room, or remotely with distributed teams.

Problem Definition Process

While problems can be complex, the problem-solving methods you use to identify and solve those problems can often be simple in design.

By taking the time to truly identify and define a problem before asking the group to reframe the challenge as an opportunity, this method is a great way to enable change.

Begin by identifying a focus question and exploring the ways in which it manifests before splitting into five teams who will each consider the problem using a different method: escape, reversal, exaggeration, distortion or wishful. Teams develop a problem objective and create ideas in line with their method before then feeding them back to the group.

This method is great for enabling in-depth discussions while also creating space for finding creative solutions too!

Problem Definition #problem solving #idea generation #creativity #online #remote-friendly A problem solving technique to define a problem, challenge or opportunity and to generate ideas.

The 5 Whys

Sometimes, a group needs to go further with their strategies and analyze the root cause at the heart of organizational issues. An RCA or root cause analysis is the process of identifying what is at the heart of business problems or recurring challenges.

The 5 Whys is a simple and effective method of helping a group go find the root cause of any problem or challenge and conduct analysis that will deliver results.

By beginning with the creation of a problem statement and going through five stages to refine it, The 5 Whys provides everything you need to truly discover the cause of an issue.

The 5 Whys #hyperisland #innovation This simple and powerful method is useful for getting to the core of a problem or challenge. As the title suggests, the group defines a problems, then asks the question “why” five times, often using the resulting explanation as a starting point for creative problem solving.

World Cafe is a simple but powerful facilitation technique to help bigger groups to focus their energy and attention on solving complex problems.

World Cafe enables this approach by creating a relaxed atmosphere where participants are able to self-organize and explore topics relevant and important to them which are themed around a central problem-solving purpose. Create the right atmosphere by modeling your space after a cafe and after guiding the group through the method, let them take the lead!

Making problem-solving a part of your organization’s culture in the long term can be a difficult undertaking. More approachable formats like World Cafe can be especially effective in bringing people unfamiliar with workshops into the fold.

World Cafe #hyperisland #innovation #issue analysis World Café is a simple yet powerful method, originated by Juanita Brown, for enabling meaningful conversations driven completely by participants and the topics that are relevant and important to them. Facilitators create a cafe-style space and provide simple guidelines. Participants then self-organize and explore a set of relevant topics or questions for conversation.

Discovery & Action Dialogue (DAD)

One of the best approaches is to create a safe space for a group to share and discover practices and behaviors that can help them find their own solutions.

With DAD, you can help a group choose which problems they wish to solve and which approaches they will take to do so. It’s great at helping remove resistance to change and can help get buy-in at every level too!

This process of enabling frontline ownership is great in ensuring follow-through and is one of the methods you will want in your toolbox as a facilitator.

Discovery & Action Dialogue (DAD) #idea generation #liberating structures #action #issue analysis #remote-friendly DADs make it easy for a group or community to discover practices and behaviors that enable some individuals (without access to special resources and facing the same constraints) to find better solutions than their peers to common problems. These are called positive deviant (PD) behaviors and practices. DADs make it possible for people in the group, unit, or community to discover by themselves these PD practices. DADs also create favorable conditions for stimulating participants’ creativity in spaces where they can feel safe to invent new and more effective practices. Resistance to change evaporates as participants are unleashed to choose freely which practices they will adopt or try and which problems they will tackle. DADs make it possible to achieve frontline ownership of solutions.

Design Sprint 2.0

Want to see how a team can solve big problems and move forward with prototyping and testing solutions in a few days? The Design Sprint 2.0 template from Jake Knapp, author of Sprint, is a complete agenda for a with proven results.

Developing the right agenda can involve difficult but necessary planning. Ensuring all the correct steps are followed can also be stressful or time-consuming depending on your level of experience.

Use this complete 4-day workshop template if you are finding there is no obvious solution to your challenge and want to focus your team around a specific problem that might require a shortcut to launching a minimum viable product or waiting for the organization-wide implementation of a solution.

Open space technology

Open space technology- developed by Harrison Owen – creates a space where large groups are invited to take ownership of their problem solving and lead individual sessions. Open space technology is a great format when you have a great deal of expertise and insight in the room and want to allow for different takes and approaches on a particular theme or problem you need to be solved.

Start by bringing your participants together to align around a central theme and focus their efforts. Explain the ground rules to help guide the problem-solving process and then invite members to identify any issue connecting to the central theme that they are interested in and are prepared to take responsibility for.

Once participants have decided on their approach to the core theme, they write their issue on a piece of paper, announce it to the group, pick a session time and place, and post the paper on the wall. As the wall fills up with sessions, the group is then invited to join the sessions that interest them the most and which they can contribute to, then you’re ready to begin!

Everyone joins the problem-solving group they’ve signed up to, record the discussion and if appropriate, findings can then be shared with the rest of the group afterward.

Open Space Technology #action plan #idea generation #problem solving #issue analysis #large group #online #remote-friendly Open Space is a methodology for large groups to create their agenda discerning important topics for discussion, suitable for conferences, community gatherings and whole system facilitation

Techniques to identify and analyze problems

Using a problem-solving method to help a team identify and analyze a problem can be a quick and effective addition to any workshop or meeting.

While further actions are always necessary, you can generate momentum and alignment easily, and these activities are a great place to get started.

We’ve put together this list of techniques to help you and your team with problem identification, analysis, and discussion that sets the foundation for developing effective solutions.

Let’s take a look!

Fishbone Analysis

Organizational or team challenges are rarely simple, and it’s important to remember that one problem can be an indication of something that goes deeper and may require further consideration to be solved.

Fishbone Analysis helps groups to dig deeper and understand the origins of a problem. It’s a great example of a root cause analysis method that is simple for everyone on a team to get their head around.

Participants in this activity are asked to annotate a diagram of a fish, first adding the problem or issue to be worked on at the head of a fish before then brainstorming the root causes of the problem and adding them as bones on the fish.

Using abstractions such as a diagram of a fish can really help a team break out of their regular thinking and develop a creative approach.

Fishbone Analysis #problem solving ##root cause analysis #decision making #online facilitation A process to help identify and understand the origins of problems, issues or observations.

Problem Tree

Encouraging visual thinking can be an essential part of many strategies. By simply reframing and clarifying problems, a group can move towards developing a problem solving model that works for them.

In Problem Tree, groups are asked to first brainstorm a list of problems – these can be design problems, team problems or larger business problems – and then organize them into a hierarchy. The hierarchy could be from most important to least important or abstract to practical, though the key thing with problem solving games that involve this aspect is that your group has some way of managing and sorting all the issues that are raised.

Once you have a list of problems that need to be solved and have organized them accordingly, you’re then well-positioned for the next problem solving steps.

Problem tree #define intentions #create #design #issue analysis A problem tree is a tool to clarify the hierarchy of problems addressed by the team within a design project; it represents high level problems or related sublevel problems.

SWOT Analysis

Chances are you’ve heard of the SWOT Analysis before. This problem-solving method focuses on identifying strengths, weaknesses, opportunities, and threats is a tried and tested method for both individuals and teams.

Start by creating a desired end state or outcome and bare this in mind – any process solving model is made more effective by knowing what you are moving towards. Create a quadrant made up of the four categories of a SWOT analysis and ask participants to generate ideas based on each of those quadrants.

Once you have those ideas assembled in their quadrants, cluster them together based on their affinity with other ideas. These clusters are then used to facilitate group conversations and move things forward.

SWOT analysis #gamestorming #problem solving #action #meeting facilitation The SWOT Analysis is a long-standing technique of looking at what we have, with respect to the desired end state, as well as what we could improve on. It gives us an opportunity to gauge approaching opportunities and dangers, and assess the seriousness of the conditions that affect our future. When we understand those conditions, we can influence what comes next.

Agreement-Certainty Matrix

Not every problem-solving approach is right for every challenge, and deciding on the right method for the challenge at hand is a key part of being an effective team.

The Agreement Certainty matrix helps teams align on the nature of the challenges facing them. By sorting problems from simple to chaotic, your team can understand what methods are suitable for each problem and what they can do to ensure effective results.

If you are already using Liberating Structures techniques as part of your problem-solving strategy, the Agreement-Certainty Matrix can be an invaluable addition to your process. We’ve found it particularly if you are having issues with recurring problems in your organization and want to go deeper in understanding the root cause.

Agreement-Certainty Matrix #issue analysis #liberating structures #problem solving You can help individuals or groups avoid the frequent mistake of trying to solve a problem with methods that are not adapted to the nature of their challenge. The combination of two questions makes it possible to easily sort challenges into four categories: simple, complicated, complex , and chaotic . A problem is simple when it can be solved reliably with practices that are easy to duplicate. It is complicated when experts are required to devise a sophisticated solution that will yield the desired results predictably. A problem is complex when there are several valid ways to proceed but outcomes are not predictable in detail. Chaotic is when the context is too turbulent to identify a path forward. A loose analogy may be used to describe these differences: simple is like following a recipe, complicated like sending a rocket to the moon, complex like raising a child, and chaotic is like the game “Pin the Tail on the Donkey.” The Liberating Structures Matching Matrix in Chapter 5 can be used as the first step to clarify the nature of a challenge and avoid the mismatches between problems and solutions that are frequently at the root of chronic, recurring problems.

Organizing and charting a team’s progress can be important in ensuring its success. SQUID (Sequential Question and Insight Diagram) is a great model that allows a team to effectively switch between giving questions and answers and develop the skills they need to stay on track throughout the process.

Begin with two different colored sticky notes – one for questions and one for answers – and with your central topic (the head of the squid) on the board. Ask the group to first come up with a series of questions connected to their best guess of how to approach the topic. Ask the group to come up with answers to those questions, fix them to the board and connect them with a line. After some discussion, go back to question mode by responding to the generated answers or other points on the board.

It’s rewarding to see a diagram grow throughout the exercise, and a completed SQUID can provide a visual resource for future effort and as an example for other teams.

SQUID #gamestorming #project planning #issue analysis #problem solving When exploring an information space, it’s important for a group to know where they are at any given time. By using SQUID, a group charts out the territory as they go and can navigate accordingly. SQUID stands for Sequential Question and Insight Diagram.

To continue with our nautical theme, Speed Boat is a short and sweet activity that can help a team quickly identify what employees, clients or service users might have a problem with and analyze what might be standing in the way of achieving a solution.

Methods that allow for a group to make observations, have insights and obtain those eureka moments quickly are invaluable when trying to solve complex problems.

In Speed Boat, the approach is to first consider what anchors and challenges might be holding an organization (or boat) back. Bonus points if you are able to identify any sharks in the water and develop ideas that can also deal with competitors!

Speed Boat #gamestorming #problem solving #action Speedboat is a short and sweet way to identify what your employees or clients don’t like about your product/service or what’s standing in the way of a desired goal.

The Journalistic Six

Some of the most effective ways of solving problems is by encouraging teams to be more inclusive and diverse in their thinking.

Based on the six key questions journalism students are taught to answer in articles and news stories, The Journalistic Six helps create teams to see the whole picture. By using who, what, when, where, why, and how to facilitate the conversation and encourage creative thinking, your team can make sure that the problem identification and problem analysis stages of the are covered exhaustively and thoughtfully. Reporter’s notebook and dictaphone optional.

The Journalistic Six – Who What When Where Why How #idea generation #issue analysis #problem solving #online #creative thinking #remote-friendly A questioning method for generating, explaining, investigating ideas.

Individual and group perspectives are incredibly important, but what happens if people are set in their minds and need a change of perspective in order to approach a problem more effectively?

Flip It is a method we love because it is both simple to understand and run, and allows groups to understand how their perspectives and biases are formed.

Participants in Flip It are first invited to consider concerns, issues, or problems from a perspective of fear and write them on a flip chart. Then, the group is asked to consider those same issues from a perspective of hope and flip their understanding.

No problem and solution is free from existing bias and by changing perspectives with Flip It, you can then develop a problem solving model quickly and effectively.

Flip It! #gamestorming #problem solving #action Often, a change in a problem or situation comes simply from a change in our perspectives. Flip It! is a quick game designed to show players that perspectives are made, not born.

LEGO Challenge

Now for an activity that is a little out of the (toy) box. LEGO Serious Play is a facilitation methodology that can be used to improve creative thinking and problem-solving skills.

The LEGO Challenge includes giving each member of the team an assignment that is hidden from the rest of the group while they create a structure without speaking.

What the LEGO challenge brings to the table is a fun working example of working with stakeholders who might not be on the same page to solve problems. Also, it’s LEGO! Who doesn’t love LEGO!

LEGO Challenge #hyperisland #team A team-building activity in which groups must work together to build a structure out of LEGO, but each individual has a secret “assignment” which makes the collaborative process more challenging. It emphasizes group communication, leadership dynamics, conflict, cooperation, patience and problem solving strategy.

What, So What, Now What?

If not carefully managed, the problem identification and problem analysis stages of the problem-solving process can actually create more problems and misunderstandings.

The What, So What, Now What? problem-solving activity is designed to help collect insights and move forward while also eliminating the possibility of disagreement when it comes to identifying, clarifying, and analyzing organizational or work problems.

Facilitation is all about bringing groups together so that might work on a shared goal and the best problem-solving strategies ensure that teams are aligned in purpose, if not initially in opinion or insight.

Throughout the three steps of this game, you give everyone on a team to reflect on a problem by asking what happened, why it is important, and what actions should then be taken.

This can be a great activity for bringing our individual perceptions about a problem or challenge and contextualizing it in a larger group setting. This is one of the most important problem-solving skills you can bring to your organization.

W³ – What, So What, Now What? #issue analysis #innovation #liberating structures You can help groups reflect on a shared experience in a way that builds understanding and spurs coordinated action while avoiding unproductive conflict. It is possible for every voice to be heard while simultaneously sifting for insights and shaping new direction. Progressing in stages makes this practical—from collecting facts about What Happened to making sense of these facts with So What and finally to what actions logically follow with Now What . The shared progression eliminates most of the misunderstandings that otherwise fuel disagreements about what to do. Voila!

Journalists

Problem analysis can be one of the most important and decisive stages of all problem-solving tools. Sometimes, a team can become bogged down in the details and are unable to move forward.

Journalists is an activity that can avoid a group from getting stuck in the problem identification or problem analysis stages of the process.

In Journalists, the group is invited to draft the front page of a fictional newspaper and figure out what stories deserve to be on the cover and what headlines those stories will have. By reframing how your problems and challenges are approached, you can help a team move productively through the process and be better prepared for the steps to follow.

Journalists #vision #big picture #issue analysis #remote-friendly This is an exercise to use when the group gets stuck in details and struggles to see the big picture. Also good for defining a vision.

Problem-solving techniques for brainstorming solutions

Now you have the context and background of the problem you are trying to solving, now comes the time to start ideating and thinking about how you’ll solve the issue.

Here, you’ll want to encourage creative, free thinking and speed. Get as many ideas out as possible and explore different perspectives so you have the raw material for the next step.

Looking at a problem from a new angle can be one of the most effective ways of creating an effective solution. TRIZ is a problem-solving tool that asks the group to consider what they must not do in order to solve a challenge.

By reversing the discussion, new topics and taboo subjects often emerge, allowing the group to think more deeply and create ideas that confront the status quo in a safe and meaningful way. If you’re working on a problem that you’ve tried to solve before, TRIZ is a great problem-solving method to help your team get unblocked.

Making Space with TRIZ #issue analysis #liberating structures #issue resolution You can clear space for innovation by helping a group let go of what it knows (but rarely admits) limits its success and by inviting creative destruction. TRIZ makes it possible to challenge sacred cows safely and encourages heretical thinking. The question “What must we stop doing to make progress on our deepest purpose?” induces seriously fun yet very courageous conversations. Since laughter often erupts, issues that are otherwise taboo get a chance to be aired and confronted. With creative destruction come opportunities for renewal as local action and innovation rush in to fill the vacuum. Whoosh!

Mindspin

Brainstorming is part of the bread and butter of the problem-solving process and all problem-solving strategies benefit from getting ideas out and challenging a team to generate solutions quickly.

With Mindspin, participants are encouraged not only to generate ideas but to do so under time constraints and by slamming down cards and passing them on. By doing multiple rounds, your team can begin with a free generation of possible solutions before moving on to developing those solutions and encouraging further ideation.

This is one of our favorite problem-solving activities and can be great for keeping the energy up throughout the workshop. Remember the importance of helping people become engaged in the process – energizing problem-solving techniques like Mindspin can help ensure your team stays engaged and happy, even when the problems they’re coming together to solve are complex.

MindSpin #teampedia #idea generation #problem solving #action A fast and loud method to enhance brainstorming within a team. Since this activity has more than round ideas that are repetitive can be ruled out leaving more creative and innovative answers to the challenge.

The Creativity Dice

One of the most useful problem solving skills you can teach your team is of approaching challenges with creativity, flexibility, and openness. Games like The Creativity Dice allow teams to overcome the potential hurdle of too much linear thinking and approach the process with a sense of fun and speed.

In The Creativity Dice, participants are organized around a topic and roll a dice to determine what they will work on for a period of 3 minutes at a time. They might roll a 3 and work on investigating factual information on the chosen topic. They might roll a 1 and work on identifying the specific goals, standards, or criteria for the session.

Encouraging rapid work and iteration while asking participants to be flexible are great skills to cultivate. Having a stage for idea incubation in this game is also important. Moments of pause can help ensure the ideas that are put forward are the most suitable.

The Creativity Dice #creativity #problem solving #thiagi #issue analysis Too much linear thinking is hazardous to creative problem solving. To be creative, you should approach the problem (or the opportunity) from different points of view. You should leave a thought hanging in mid-air and move to another. This skipping around prevents premature closure and lets your brain incubate one line of thought while you consciously pursue another.

Idea and Concept Development

Brainstorming without structure can quickly become chaotic or frustrating. In a problem-solving context, having an ideation framework to follow can help ensure your team is both creative and disciplined.

In this method, you’ll find an idea generation process that encourages your group to brainstorm effectively before developing their ideas and begin clustering them together. By using concepts such as Yes and…, more is more and postponing judgement, you can create the ideal conditions for brainstorming with ease.

Idea & Concept Development #hyperisland #innovation #idea generation Ideation and Concept Development is a process for groups to work creatively and collaboratively to generate creative ideas. It’s a general approach that can be adapted and customized to suit many different scenarios. It includes basic principles for idea generation and several steps for groups to work with. It also includes steps for idea selection and development.

Problem-solving techniques for developing and refining solutions

The success of any problem-solving process can be measured by the solutions it produces. After you’ve defined the issue, explored existing ideas, and ideated, it’s time to develop and refine your ideas in order to bring them closer to a solution that actually solves the problem.

Use these problem-solving techniques when you want to help your team think through their ideas and refine them as part of your problem solving process.

Improved Solutions

After a team has successfully identified a problem and come up with a few solutions, it can be tempting to call the work of the problem-solving process complete. That said, the first solution is not necessarily the best, and by including a further review and reflection activity into your problem-solving model, you can ensure your group reaches the best possible result.

One of a number of problem-solving games from Thiagi Group, Improved Solutions helps you go the extra mile and develop suggested solutions with close consideration and peer review. By supporting the discussion of several problems at once and by shifting team roles throughout, this problem-solving technique is a dynamic way of finding the best solution.

Improved Solutions #creativity #thiagi #problem solving #action #team You can improve any solution by objectively reviewing its strengths and weaknesses and making suitable adjustments. In this creativity framegame, you improve the solutions to several problems. To maintain objective detachment, you deal with a different problem during each of six rounds and assume different roles (problem owner, consultant, basher, booster, enhancer, and evaluator) during each round. At the conclusion of the activity, each player ends up with two solutions to her problem.

Four Step Sketch

Creative thinking and visual ideation does not need to be confined to the opening stages of your problem-solving strategies. Exercises that include sketching and prototyping on paper can be effective at the solution finding and development stage of the process, and can be great for keeping a team engaged.

By going from simple notes to a crazy 8s round that involves rapidly sketching 8 variations on their ideas before then producing a final solution sketch, the group is able to iterate quickly and visually. Problem-solving techniques like Four-Step Sketch are great if you have a group of different thinkers and want to change things up from a more textual or discussion-based approach.

Four-Step Sketch #design sprint #innovation #idea generation #remote-friendly The four-step sketch is an exercise that helps people to create well-formed concepts through a structured process that includes: Review key information Start design work on paper, Consider multiple variations , Create a detailed solution . This exercise is preceded by a set of other activities allowing the group to clarify the challenge they want to solve. See how the Four Step Sketch exercise fits into a Design Sprint

Ensuring that everyone in a group is able to contribute to a discussion is vital during any problem solving process. Not only does this ensure all bases are covered, but its then easier to get buy-in and accountability when people have been able to contribute to the process.

1-2-4-All is a tried and tested facilitation technique where participants are asked to first brainstorm on a topic on their own. Next, they discuss and share ideas in a pair before moving into a small group. Those groups are then asked to present the best idea from their discussion to the rest of the team.

This method can be used in many different contexts effectively, though I find it particularly shines in the idea development stage of the process. Giving each participant time to concretize their ideas and develop them in progressively larger groups can create a great space for both innovation and psychological safety.

1-2-4-All #idea generation #liberating structures #issue analysis With this facilitation technique you can immediately include everyone regardless of how large the group is. You can generate better ideas and more of them faster than ever before. You can tap the know-how and imagination that is distributed widely in places not known in advance. Open, generative conversation unfolds. Ideas and solutions are sifted in rapid fashion. Most importantly, participants own the ideas, so follow-up and implementation is simplified. No buy-in strategies needed! Simple and elegant!

15% Solutions

Some problems are simpler than others and with the right problem-solving activities, you can empower people to take immediate actions that can help create organizational change.

Part of the liberating structures toolkit, 15% solutions is a problem-solving technique that focuses on finding and implementing solutions quickly. A process of iterating and making small changes quickly can help generate momentum and an appetite for solving complex problems.

Problem-solving strategies can live and die on whether people are onboard. Getting some quick wins is a great way of getting people behind the process.

It can be extremely empowering for a team to realize that problem-solving techniques can be deployed quickly and easily and delineate between things they can positively impact and those things they cannot change.

15% Solutions #action #liberating structures #remote-friendly You can reveal the actions, however small, that everyone can do immediately. At a minimum, these will create momentum, and that may make a BIG difference. 15% Solutions show that there is no reason to wait around, feel powerless, or fearful. They help people pick it up a level. They get individuals and the group to focus on what is within their discretion instead of what they cannot change. With a very simple question, you can flip the conversation to what can be done and find solutions to big problems that are often distributed widely in places not known in advance. Shifting a few grains of sand may trigger a landslide and change the whole landscape.

Problem-solving techniques for making decisions and planning

After your group is happy with the possible solutions you’ve developed, now comes the time to choose which to implement. There’s more than one way to make a decision and the best option is often dependant on the needs and set-up of your group.

Sometimes, it’s the case that you’ll want to vote as a group on what is likely to be the most impactful solution. Other times, it might be down to a decision maker or major stakeholder to make the final decision. Whatever your process, here’s some techniques you can use to help you make a decision during your problem solving process.

How-Now-Wow Matrix

The problem-solving process is often creative, as complex problems usually require a change of thinking and creative response in order to find the best solutions. While it’s common for the first stages to encourage creative thinking, groups can often gravitate to familiar solutions when it comes to the end of the process.

When selecting solutions, you don’t want to lose your creative energy! The How-Now-Wow Matrix from Gamestorming is a great problem-solving activity that enables a group to stay creative and think out of the box when it comes to selecting the right solution for a given problem.

Problem-solving techniques that encourage creative thinking and the ideation and selection of new solutions can be the most effective in organisational change. Give the How-Now-Wow Matrix a go, and not just for how pleasant it is to say out loud.

How-Now-Wow Matrix #gamestorming #idea generation #remote-friendly When people want to develop new ideas, they most often think out of the box in the brainstorming or divergent phase. However, when it comes to convergence, people often end up picking ideas that are most familiar to them. This is called a ‘creative paradox’ or a ‘creadox’. The How-Now-Wow matrix is an idea selection tool that breaks the creadox by forcing people to weigh each idea on 2 parameters.

Impact and Effort Matrix

All problem-solving techniques hope to not only find solutions to a given problem or challenge but to find the best solution. When it comes to finding a solution, groups are invited to put on their decision-making hats and really think about how a proposed idea would work in practice.

The Impact and Effort Matrix is one of the problem-solving techniques that fall into this camp, empowering participants to first generate ideas and then categorize them into a 2×2 matrix based on impact and effort.

Activities that invite critical thinking while remaining simple are invaluable. Use the Impact and Effort Matrix to move from ideation and towards evaluating potential solutions before then committing to them.

Impact and Effort Matrix #gamestorming #decision making #action #remote-friendly In this decision-making exercise, possible actions are mapped based on two factors: effort required to implement and potential impact. Categorizing ideas along these lines is a useful technique in decision making, as it obliges contributors to balance and evaluate suggested actions before committing to them.

If you’ve followed each of the problem-solving steps with your group successfully, you should move towards the end of your process with heaps of possible solutions developed with a specific problem in mind. But how do you help a group go from ideation to putting a solution into action?

Dotmocracy – or Dot Voting -is a tried and tested method of helping a team in the problem-solving process make decisions and put actions in place with a degree of oversight and consensus.

One of the problem-solving techniques that should be in every facilitator’s toolbox, Dot Voting is fast and effective and can help identify the most popular and best solutions and help bring a group to a decision effectively.

Dotmocracy #action #decision making #group prioritization #hyperisland #remote-friendly Dotmocracy is a simple method for group prioritization or decision-making. It is not an activity on its own, but a method to use in processes where prioritization or decision-making is the aim. The method supports a group to quickly see which options are most popular or relevant. The options or ideas are written on post-its and stuck up on a wall for the whole group to see. Each person votes for the options they think are the strongest, and that information is used to inform a decision.

Straddling the gap between decision making and planning, MoSCoW is a simple and effective method that allows a group team to easily prioritize a set of possible options.

Use this method in a problem solving process by collecting and summarizing all your possible solutions and then categorize them into 4 sections: “Must have”, “Should have”, “Could have”, or “Would like but won‘t get”.

This method is particularly useful when its less about choosing one possible solution and more about prioritorizing which to do first and which may not fit in the scope of your project. In my experience, complex challenges often require multiple small fixes, and this method can be a great way to move from a pile of things you’d all like to do to a structured plan.

MoSCoW #define intentions #create #design #action #remote-friendly MoSCoW is a method that allows the team to prioritize the different features that they will work on. Features are then categorized into “Must have”, “Should have”, “Could have”, or “Would like but won‘t get”. To be used at the beginning of a timeslot (for example during Sprint planning) and when planning is needed.

When it comes to managing the rollout of a solution, clarity and accountability are key factors in ensuring the success of the project. The RAACI chart is a simple but effective model for setting roles and responsibilities as part of a planning session.

Start by listing each person involved in the project and put them into the following groups in order to make it clear who is responsible for what during the rollout of your solution.

Responsibility (Which person and/or team will be taking action?)
Authority (At what “point” must the responsible person check in before going further?)
Accountability (Who must the responsible person check in with?)
Consultation (Who must be consulted by the responsible person before decisions are made?)
Information (Who must be informed of decisions, once made?)

Ensure this information is easily accessible and use it to inform who does what and who is looped into discussions and kept up to date.

RAACI #roles and responsibility #teamwork #project management Clarifying roles and responsibilities, levels of autonomy/latitude in decision making, and levels of engagement among diverse stakeholders.

Problem-solving warm-up activities

All facilitators know that warm-ups and icebreakers are useful for any workshop or group process. Problem-solving workshops are no different.

Use these problem-solving techniques to warm up a group and prepare them for the rest of the process. Activating your group by tapping into some of the top problem-solving skills can be one of the best ways to see great outcomes from your session.

Check-in / Check-out

Solid processes are planned from beginning to end, and the best facilitators know that setting the tone and establishing a safe, open environment can be integral to a successful problem-solving process. Check-in / Check-out is a great way to begin and/or bookend a problem-solving workshop. Checking in to a session emphasizes that everyone will be seen, heard, and expected to contribute.

If you are running a series of meetings, setting a consistent pattern of checking in and checking out can really help your team get into a groove. We recommend this opening-closing activity for small to medium-sized groups though it can work with large groups if they’re disciplined!

Check-in / Check-out #team #opening #closing #hyperisland #remote-friendly Either checking-in or checking-out is a simple way for a team to open or close a process, symbolically and in a collaborative way. Checking-in/out invites each member in a group to be present, seen and heard, and to express a reflection or a feeling. Checking-in emphasizes presence, focus and group commitment; checking-out emphasizes reflection and symbolic closure.

Doodling Together

Thinking creatively and not being afraid to make suggestions are important problem-solving skills for any group or team, and warming up by encouraging these behaviors is a great way to start.

Doodling Together is one of our favorite creative ice breaker games – it’s quick, effective, and fun and can make all following problem-solving steps easier by encouraging a group to collaborate visually. By passing cards and adding additional items as they go, the workshop group gets into a groove of co-creation and idea development that is crucial to finding solutions to problems.

Doodling Together #collaboration #creativity #teamwork #fun #team #visual methods #energiser #icebreaker #remote-friendly Create wild, weird and often funny postcards together & establish a group’s creative confidence.

Show and Tell

You might remember some version of Show and Tell from being a kid in school and it’s a great problem-solving activity to kick off a session.

Asking participants to prepare a little something before a workshop by bringing an object for show and tell can help them warm up before the session has even begun! Games that include a physical object can also help encourage early engagement before moving onto more big-picture thinking.

By asking your participants to tell stories about why they chose to bring a particular item to the group, you can help teams see things from new perspectives and see both differences and similarities in the way they approach a topic. Great groundwork for approaching a problem-solving process as a team!

Show and Tell #gamestorming #action #opening #meeting facilitation Show and Tell taps into the power of metaphors to reveal players’ underlying assumptions and associations around a topic The aim of the game is to get a deeper understanding of stakeholders’ perspectives on anything—a new project, an organizational restructuring, a shift in the company’s vision or team dynamic.

Constellations

Who doesn’t love stars? Constellations is a great warm-up activity for any workshop as it gets people up off their feet, energized, and ready to engage in new ways with established topics. It’s also great for showing existing beliefs, biases, and patterns that can come into play as part of your session.

Using warm-up games that help build trust and connection while also allowing for non-verbal responses can be great for easing people into the problem-solving process and encouraging engagement from everyone in the group. Constellations is great in large spaces that allow for movement and is definitely a practical exercise to allow the group to see patterns that are otherwise invisible.

Constellations #trust #connection #opening #coaching #patterns #system Individuals express their response to a statement or idea by standing closer or further from a central object. Used with teams to reveal system, hidden patterns, perspectives.

Draw a Tree

Problem-solving games that help raise group awareness through a central, unifying metaphor can be effective ways to warm-up a group in any problem-solving model.

Draw a Tree is a simple warm-up activity you can use in any group and which can provide a quick jolt of energy. Start by asking your participants to draw a tree in just 45 seconds – they can choose whether it will be abstract or realistic.

Once the timer is up, ask the group how many people included the roots of the tree and use this as a means to discuss how we can ignore important parts of any system simply because they are not visible.

All problem-solving strategies are made more effective by thinking of problems critically and by exposing things that may not normally come to light. Warm-up games like Draw a Tree are great in that they quickly demonstrate some key problem-solving skills in an accessible and effective way.

Draw a Tree #thiagi #opening #perspectives #remote-friendly With this game you can raise awarness about being more mindful, and aware of the environment we live in.

Closing activities for a problem-solving process

Each step of the problem-solving workshop benefits from an intelligent deployment of activities, games, and techniques. Bringing your session to an effective close helps ensure that solutions are followed through on and that you also celebrate what has been achieved.

Here are some problem-solving activities you can use to effectively close a workshop or meeting and ensure the great work you’ve done can continue afterward.

One Breath Feedback

Maintaining attention and focus during the closing stages of a problem-solving workshop can be tricky and so being concise when giving feedback can be important. It’s easy to incur “death by feedback” should some team members go on for too long sharing their perspectives in a quick feedback round.

One Breath Feedback is a great closing activity for workshops. You give everyone an opportunity to provide feedback on what they’ve done but only in the space of a single breath. This keeps feedback short and to the point and means that everyone is encouraged to provide the most important piece of feedback to them.

One breath feedback #closing #feedback #action This is a feedback round in just one breath that excels in maintaining attention: each participants is able to speak during just one breath … for most people that’s around 20 to 25 seconds … unless of course you’ve been a deep sea diver in which case you’ll be able to do it for longer.

Who What When Matrix

Matrices feature as part of many effective problem-solving strategies and with good reason. They are easily recognizable, simple to use, and generate results.

The Who What When Matrix is a great tool to use when closing your problem-solving session by attributing a who, what and when to the actions and solutions you have decided upon. The resulting matrix is a simple, easy-to-follow way of ensuring your team can move forward.

Great solutions can’t be enacted without action and ownership. Your problem-solving process should include a stage for allocating tasks to individuals or teams and creating a realistic timeframe for those solutions to be implemented or checked out. Use this method to keep the solution implementation process clear and simple for all involved.

Who/What/When Matrix #gamestorming #action #project planning With Who/What/When matrix, you can connect people with clear actions they have defined and have committed to.

Response cards

Group discussion can comprise the bulk of most problem-solving activities and by the end of the process, you might find that your team is talked out!

Providing a means for your team to give feedback with short written notes can ensure everyone is head and can contribute without the need to stand up and talk. Depending on the needs of the group, giving an alternative can help ensure everyone can contribute to your problem-solving model in the way that makes the most sense for them.

Response Cards is a great way to close a workshop if you are looking for a gentle warm-down and want to get some swift discussion around some of the feedback that is raised.

Response Cards #debriefing #closing #structured sharing #questions and answers #thiagi #action It can be hard to involve everyone during a closing of a session. Some might stay in the background or get unheard because of louder participants. However, with the use of Response Cards, everyone will be involved in providing feedback or clarify questions at the end of a session.

Tips for effective problem solving

Problem-solving activities are only one part of the puzzle. While a great method can help unlock your team’s ability to solve problems, without a thoughtful approach and strong facilitation the solutions may not be fit for purpose.

Let’s take a look at some problem-solving tips you can apply to any process to help it be a success!

Clearly define the problem

Jumping straight to solutions can be tempting, though without first clearly articulating a problem, the solution might not be the right one. Many of the problem-solving activities below include sections where the problem is explored and clearly defined before moving on.

This is a vital part of the problem-solving process and taking the time to fully define an issue can save time and effort later. A clear definition helps identify irrelevant information and it also ensures that your team sets off on the right track.

Don’t jump to conclusions

It’s easy for groups to exhibit cognitive bias or have preconceived ideas about both problems and potential solutions. Be sure to back up any problem statements or potential solutions with facts, research, and adequate forethought.

The best techniques ask participants to be methodical and challenge preconceived notions. Make sure you give the group enough time and space to collect relevant information and consider the problem in a new way. By approaching the process with a clear, rational mindset, you’ll often find that better solutions are more forthcoming.

Try different approaches

Problems come in all shapes and sizes and so too should the methods you use to solve them. If you find that one approach isn’t yielding results and your team isn’t finding different solutions, try mixing it up. You’ll be surprised at how using a new creative activity can unblock your team and generate great solutions.

Don’t take it personally

Depending on the nature of your team or organizational problems, it’s easy for conversations to get heated. While it’s good for participants to be engaged in the discussions, ensure that emotions don’t run too high and that blame isn’t thrown around while finding solutions.

You’re all in it together, and even if your team or area is seeing problems, that isn’t necessarily a disparagement of you personally. Using facilitation skills to manage group dynamics is one effective method of helping conversations be more constructive.

Get the right people in the room

Your problem-solving method is often only as effective as the group using it. Getting the right people on the job and managing the number of people present is important too!

If the group is too small, you may not get enough different perspectives to effectively solve a problem. If the group is too large, you can go round and round during the ideation stages.

Creating the right group makeup is also important in ensuring you have the necessary expertise and skillset to both identify and follow up on potential solutions. Carefully consider who to include at each stage to help ensure your problem-solving method is followed and positioned for success.

Create psychologically safe spaces for discussion

Identifying a problem accurately also requires that all members of a group are able to contribute their views in an open and safe manner.

It can be tough for people to stand up and contribute if the problems or challenges are emotive or personal in nature. Try and create a psychologically safe space for these kinds of discussions and where possible, create regular opportunities for challenges to be brought up organically.

Document everything

The best solutions can take refinement, iteration, and reflection to come out. Get into a habit of documenting your process in order to keep all the learnings from the session and to allow ideas to mature and develop. Many of the methods below involve the creation of documents or shared resources. Be sure to keep and share these so everyone can benefit from the work done!

Bring a facilitator

Facilitation is all about making group processes easier. With a subject as potentially emotive and important as problem-solving, having an impartial third party in the form of a facilitator can make all the difference in finding great solutions and keeping the process moving. Consider bringing a facilitator to your problem-solving session to get better results and generate meaningful solutions!

Develop your problem-solving skills

It takes time and practice to be an effective problem solver. While some roles or participants might more naturally gravitate towards problem-solving, it can take development and planning to help everyone create better solutions.

You might develop a training program, run a problem-solving workshop or simply ask your team to practice using the techniques below. Check out our post on problem-solving skills to see how you and your group can develop the right mental process and be more resilient to issues too!

Design a great agenda

Workshops are a great format for solving problems. With the right approach, you can focus a group and help them find the solutions to their own problems. But designing a process can be time-consuming and finding the right activities can be difficult.

Check out our workshop planning guide to level-up your agenda design and start running more effective workshops. Need inspiration? Check out templates designed by expert facilitators to help you kickstart your process!

Save time and effort creating an effective problem solving process

A structured problem solving process is a surefire way of solving tough problems, discovering creative solutions and driving organizational change. But how can you design for successful outcomes?

With SessionLab, it’s easy to design engaging workshops that deliver results. Drag, drop and reorder blocks to build your agenda. When you make changes or update your agenda, your session timing adjusts automatically , saving you time on manual adjustments.

Collaborating with stakeholders or clients? Share your agenda with a single click and collaborate in real-time. No more sending documents back and forth over email.

Explore how to use SessionLab to design effective problem solving workshops or watch this five minute video to see the planner in action!

Over to you

The problem-solving process can often be as complicated and multifaceted as the problems they are set-up to solve. With the right problem-solving techniques and a mix of exercises designed to guide discussion and generate purposeful ideas, we hope we’ve given you the tools to find the best solutions as simply and easily as possible.

Is there a problem-solving technique that you are missing here? Do you have a favorite activity or method you use when facilitating? Let us know in the comments below, we’d love to hear from you!

James Smart is Head of Content at SessionLab. He’s also a creative facilitator who has run workshops and designed courses for establishments like the National Centre for Writing, UK. He especially enjoys working with young people and empowering others in their creative practice.

thank you very much for these excellent techniques

Certainly wonderful article, very detailed. Shared!

Your list of techniques for problem solving can be helpfully extended by adding TRIZ to the list of techniques. TRIZ has 40 problem solving techniques derived from methods inventros and patent holders used to get new patents. About 10-12 are general approaches. many organization sponsor classes in TRIZ that are used to solve business problems or general organiztational problems. You can take a look at TRIZ and dwonload a free internet booklet to see if you feel it shound be included per your selection process.

Design your next workshop with SessionLab

Join the 150,000 facilitators using SessionLab

Time and Work

In time and work we will learn to calculate and find the time required to complete a piece of work and also find work done in a given period of time. We know the amount of work done by a person varies directly with the time taken by him to complete the work. (i) Suppose A can finish a piece of work in 8 days. Then, work done by A in 1 day = ¹/₈ [by unitary method]. (ii) Suppose that the work done by A in 1 day is ¹/₆ Then, time taken by A to finish the whole work = 6 days.

General Rules

(i) Suppose if a person A can finish a work in n days. Then, work done by A in 1 day = 1/nᵗʰ part of the work.

(ii) Suppose that the work done by A in 1 day is $\frac{1}{n}$ Then, time taken by A to finish the whole work = n days.

Problems on Time and Work :

1. aaron alone can finish a piece of work in 12 days and brandon alone can do it in 15 days. if both of them work at it together, how much time will they take to finish it.

Solution: Time taken by Aaron to finish the work = 12 days. Work done by Aaron in 1 day = ¹/₁₂

Time taken by Brandon to finish the work = 15 days. Work done by Brandon in 1 day = ¹/₁₅ Work done by (Aaron + Brandon) in 1 day = ¹/₁₂ + ¹/₁₅ = ⁹/₆₀ = ³/₂₀ Time taken by (Aaron + Brandon) to finish the work = $\frac{20}{6}$ days, i.e., 6²/₃ days.

Hence both can finish the work in 6²/₃ days.

2. A and B together can do a piece of work in 15 days, while B alone can finish it 20 days. In how many days can A alone finish the work?

Solution: Time taken by (A + B) to finish the work = 15 days. Time taken by B alone to finish the work 20 days. (A + B)’s 1 day’s work = ¹/₁₅ and B’s 1 day’s work = ¹/₂₀ A’s 1 day’s work = {(A + B)’s 1 day’s work} - {B’s 1 day’s work} = (¹/₁₅ - ¹/₂₀) = (4 - 3)/60 = ¹/₆₀ Therefore, A alone can finish the work in 60 days.

3. A can do a piece of work in 25 days and B can finish it in 20 days. They work together for 5 days and then A leaves. In how many days will B finish the remaining work?

4. a and b can do a piece of work in 18 days; b and c can do it in 24 days while c and a can finish it in 36 days. if a, b, c works together, in how many days will they finish the work.

Solution: Time taken by (A + B) to finish the work = 18 days. (A + B)’s 1 day’s work = ¹/₁₈ Time taken by (B + C) to finish the work = 24 days. (B + C)’s 1 day’s work = ¹/₂₄ Time taken by (C + A) to finish the work = 36 days. (C + A)’s 1 day’s work = ¹/₃₆ Therefore, 2(A + B + C)’s 1 day’s work = (¹/₁₈ + ¹/₂₄ + ¹/₃₆) = (4 + 3 + 2)/72 = $\frac{9}{72}$ = ¹/₈

⇒ (A + B + C)’s 1 day’s work = (¹/₂ × ¹/₈) = ¹/₁₆ Therefore, A, B, C together can finish the work in 16 days.

5. A and B can do a piece of work in 12 days; B and C can do it in 15 days while C and A can finish it in 20 days. If A, B, C works together, in how many days will they finish the work? In how many days will each one of them finish it, working alone?

Solution: Time taken by (A + B) to finish the work = 12 days. (A + B)’s 1 day’s work = ¹/₁₂ Time taken by (B +C) to finish the work = 15 days. (B + C)’s 1 day’s work = ¹/₁₅ Time taken by (C + A) to finish the work = 20 days. (C + A)’s 1 day’s work = ¹/₂₀ Therefore, 2(A + B + C)’s 1 day’s work = (¹/₁₂ + ¹/₁₅ + ¹/₂₀) = $\frac{12}{60}$ = ¹/₅

⇒ (A + B + C)’s 1 day’s work = (¹/₂ × ¹/₅) = ¹/₁₀ Therefore, A, B, C together can finish the work in 10 days. Now, A’s 1 day’s work = {(A + B + C)’s 1 day’s work} - {(B + C)’s 1 day’s work} = (¹/₁₀ - ¹/₁₅) = ¹/₃₀ Hence, A alone can finish the work in 30 days. B’s 1 day’s work {(A + B + C)’s 1 day’s work} - {(C + A)’s 1 day’s work} (¹/₁₀ – ¹/₂₀) = ¹/₂₀ Hence, B alone can finish the work in 20 days. C’s 1 days work = {(A + B + C)’s 1 day’s work} - {(A + B)’s 1 day’s work} = (¹/₁₀ – ¹/₁₂) = ¹/₆₀ Hence, C alone can finish the work in 60 days.

● Time and Work

Pipes and Cistern

Practice Test on Time and Work

● Time and Work - Worksheets

Worksheet on Time and Work

8th Grade Math Practice From Time and Work to HOME PAGE

Didn't find what you were looking for? Or want to know more information about Math Only Math . Use this Google Search to find what you need.

New! Comments

Share this page: What’s this?

Preschool Activities
Kindergarten Math
1st Grade Math
2nd Grade Math
3rd Grade Math
4th Grade Math
5th Grade Math
6th Grade Math
7th Grade Math
8th Grade Math
9th Grade Math
10th Grade Math
11 & 12 Grade Math
Concepts of Sets
Probability
Boolean Algebra
Math Coloring Pages
Multiplication Table
Cool Maths Games
Math Flash Cards
Online Math Quiz
Math Puzzles
Binary System
Math Dictionary
Conversion Chart
Homework Sheets
Math Problem Ans
Free Math Answers
Printable Math Sheet
Funny Math Answers
Employment Test
Math Patterns
Link Partners
Privacy Policy

E-mail Address

First Name

to send you Math Only Math.

Recent Articles

Worksheet on bar graphs | bar graphs or column graphs | graphing bar.

Sep 04, 24 03:48 PM

1st Grade Numbers Worksheet | Before, After and Between | Compare Numb

Sep 03, 24 03:33 PM

$1st Grade Color by Numbers$

Subtraction Word Problems - 2-Digit Numbers | Subtraction Problems

Sep 03, 24 02:26 AM

Days of the Week | 7 Days of the Week | What are the Seven Days?

Sep 03, 24 01:58 AM

$Days of the Weeks$

Time Duration |How to Calculate the Time Duration (in Hours & Minutes)

Sep 03, 24 01:37 AM

$Duration of Time$

Talk to our experts

1800-120-456-456

Work Done Problems

A Brief Explanation of Work Done Problems

Work problems typically occur when two people are painting a house together. You are typically asked how long it takes each individual to paint a house of a comparable size and how long it will take the two of them to paint the house when they collaborate. So, here the concept of work done will be used. In this article, we will learn how we can solve the work done problems.

Work Done Related Important Formula

In time and work, we will learn to calculate and determine the number of hours needed to complete a task as well as the amount of work completed in a specific period of time. We are aware that a person's productivity is closely correlated with the time it takes him to do a task.

For Example: Suppose Shyam can finish a work in 7 days.

Then, work done by Shyam in 1 day $=\dfrac{1}{7}$

Suppose if a person A can finish a work in $\mathrm{n}$ days.

Then, work done by $A$ in 1 day $=1 / n^{\text {th }}$ part of the work.

Suppose that the work done by $\mathrm{A}$ in 1 day is $\dfrac{1}{n}$

Then, time taken by $\mathrm{A}$ to finish the whole work $=\mathrm{n}$ days.

Solved Problems on Work Done

Here are some solved problems on work done , through which it can be understood in a better way:

Q1. Piyush and Rahul together can complete a work in 18 days. Piyush alone can do the same work in 24 days. What will be the number of days Rahul alone can complete the whole work?

Ans: Piyush and Rahul can complete the work in 18 days

Piyush alone can complete the work in 24 days

Taking the L.C.M of 18 and 24

L.C.M of 18 and 24 is 72

$\Rightarrow$ One day work of Piyush and Rahul $=\dfrac{72}{18}=4$

$\Rightarrow$ One day work of Piyush $=\dfrac{72}{24}=3$

$\Rightarrow$ One day work of Rahul $=4-3=1$

$\Rightarrow$ Number of days Rahul alone takes to complete the work $=\dfrac{72}{1}=72$

$\therefore$ The number of days Rahul takes to complete the whole work is 72.

Q2. A and $B$ together can do a piece of work in 15 days, while $B$ alone can finish it in 20 days. In how many days can $A$ alone finish the work?

Ans: Time taken by $(A+B)$ to finish the work $=15$ days.

Time taken by B alone to finish the work is 20 days.

$(A+B)$ 's 1 day's work $=\dfrac{1}{15}$

and $B^{\prime}$ s 1 day's work $=\dfrac{1}{20}$

A's 1 day's work $=\left\{(A+B)^{\prime}\right.$ s 1 day's work $\}-\left\{B^{\prime}\right.$ s 1 day's work $\}$

$=(\dfrac{1}{15}-\dfrac{1}{20})=\dfrac{4-3}{60}=\dfrac{1}{60}$

Therefore, A alone can finish the work in 60 days.

Q 3. A can do a piece of work in 25 days and $B$ can finish it in 20 days. They work together for 5 days and then A leaves. In how many days will $B$ finish the remaining work?

Ans: Time taken by $\mathrm{A}$ to finish the work $=25$ days.

A's 1 day's work $=\dfrac{1}{25}$

Time taken by $B$ to finish the work $=20$ days.

B's 1 day's work $=\dfrac{1}{20}$

$(A+B)$ 's 1 day's work $=(\dfrac{1}{25}+\dfrac{1}{20})=\dfrac{9}{100}$

$(A+B)$ 's 5 day's work $(5 \times \dfrac{9}{100})=\dfrac{45}{100}=\dfrac{9}{20}$

Remaining work $(1-\dfrac{9}{20})=\dfrac{11}{20}$

Now, $\dfrac{11}{20}$ work is done by $B$ in 1 day

Therefore, $\dfrac{11}{20}$ work will be done by $B$ in $(\dfrac{11}{20} \times 20)$ days $=11$ days.

Hence, the remaining work is done by $B$ in 11 days.

Practice Questions

Here are practice questions related to work done, through which it can be made in a better way:

Q1. In 300 days, Sanjay finished the school project. If Piyush is 50% more productive than Sanjay, how many days will it take him to finish the identical task?

Q2. A task can be completed by Sourav and Anshu in 18 days. Anshu and Himanshu can do it in 24 and Sourav and Himanshu can do it in 36 days, respectively. How many days will it take Sourav, Himanshu, and Anshu to complete the task if they collaborate?

Ans . 16 Days

Q3. In 600 days, Sanjay finished the school project. If Piyush is 20% more productive than Sanjay, how many days will it take him to finish the identical task?

Q4. Piyush, Santosh, and Ramesh are hired as construction workers by a builder on one of his projects. They finish a piece of work in 20, 30, and 60 days, respectively. If Santosh and Ramesh help Piyush every third day, how many days will it take him to do the entire task?

Ans . 15 Days

Q5. A project that Santosh and Prajapati are working on can be finished in 30 days. Santosh put in 16 days of labour, and Prajapati took 44 days to finish it all. How many days would it have taken Prajapati to do the entire project on her own?

Ans . 60 Days

In this article, we discussed the topic of time and work. Time and work are related concepts. Time is a unit of time, work is an activity done in a given time. Time and work are very important in the field of mathematics. This is because it helps them understand the concept of time and how it can be used to solve mathematical equations. The amount of work you do is related to the amount of time you spend on it. We have understood the topic of time and work perfectly by using some solved problems on work done and time spent.

FAQs on Work Done Problems

1. What does the term "word problems" mean?

Word problems are defined as mathematical problems that are written in an ordinary language rather than mathematical terms and symbols.

2. What are some of the different work done problems?

Problems based on Painting and Pipes, Tubs and Man-Hours, Unequal Times Etc. are some of the different work done problems.

3. Are time and work directly or inversely correlated?

The relationship between time and work is directly correlated. As a result, more work may be done in more time, and vice versa.

Testimonial
Web Stories

Learning Home

Not Now! Will rate later

Time and Work Formula and Solved Problems

The basic formula for solving is: 1/r + 1/s = 1/h
Let us take a case, say a person Hrithik
Let us say that in 1 day Hrithik will do 1/20 th of the work and 1 day Dhoni will do 1/30 th of the work. Now if they are working together they will be doing 1/20 + 1/30 = 5/60 = 1/12 th of the work in 1 day. Now try to analyze, if two persons are doing 1/12 th of the work on first day, they will do 1/12 th of the work on second day, 1/12 th of the work on third day and so on. Now adding all that when they would have worked for 12 days 12/12 = 1 i.e. the whole work would have been over. Thus the concept works in direct as well as in reverse condition.
The conclusion of the concept is if a person does a work in ‘r’ days, then in 1 day- 1/r th of the work is done and if 1/s th of the work is done in 1 day, then the work will be finished in ‘s’ days. Thus working together both can finish 1/h (1/r + 1/s = 1/h) work in 1 day & this complete the task in ’h’ hours.
The same can also be interpreted in another manner i.e. If one person does a piece of work in x days and another person does it in y days. Then together they can finish that work in xy/(x+y) days
In case of three persons taking x, y and z days respectively, They can finish the work together in xyz/(xy + yz + xz) days

Time and work problems

Time and work concepts, time and work problems (easy), time and work problems (difficult).

Most Popular Articles - PS

Problems on Ages Practice Problems : Level 02

Chain Rule : Theory & Concepts

Chain Rule Solved Examples

Chain Rule Practice Problems: Level 01

Chain Rule Practice Problems : Level 02

Problems on Numbers System : Level 02

Download our app.

Learn on-the-go
Unlimited Prep Resources
Better Learning Experience
Personalized Guidance

Get More Out of Your Exam Preparation - Try Our App!

Math Work Problems - Two Persons

In these lessons, we will learn how to solve work problems that involve two persons who may work at different rates.

Related Pages Work Problems Solving Work Word Problems Using Algebra More Algebra Lessons

Work Problems are word problems that involve different people doing work together but at different rates . If the people were working at the same rate then we can use the Inversely Proportional Method instead.

How To Solve Work Problems: Two Persons, Unknown Time

We will learn how to solve math work problems that involve two persons. We will also learn how to solve work problems with unknown time.

The following diagram shows the formula for Work Problems that involve two persons. Scroll down the page for more examples and solutions on solving algebra work problems.

This formula can be extended for more than two persons .

"Work" Problems: Two Persons

Example: Peter can mow the lawn in 40 minutes and John can mow the lawn in 60 minutes. How long will it take for them to mow the lawn together?

Solution: Step 1: Assign variables : Let x = time to mow lawn together.

Step 3: Solve the equation The LCM of 40 and 60 is 120 Multiply both sides with 120

Answer: The time taken for both of them to mow the lawn together is 24 minutes.

Work Problems With One Unknown Time

Catherine can paint a house in 15 hours. Dan can paint a house in 30 hours. How long will it take them working together.
Evan can clean a room in 3 hours. If his sister, Faith helps, it takes them two and two-fifths hours. How long will it take Faith working alone?

Variations Of GMAT Combined Work Problems

Working at a constant rate, Joe can paint a fence in 4 hours. Working at a constant rate, his brother can paint the same fence in 2 hours. How long will it take them to paint the fence if they both work together at their respective constant rates?
Working alone at a constant rate, machine A takes 2 hours to build a care. Working alone at a constant rate, machine B takes 3 hours to build the same car. If they work together for 1 hour at their respective constant rates and then machine B breaks down, how much additional time will it take machine A to finish the car by itself?
Working alone at a constant rate, Carla can wash a load of dishes in 42 minutes. If Carla works together with Dan and they both work at constant rates, it takes them 28 minutes to wash the same load of dishes. Working at a constant rate, how long would it take Dan to wash the load of dishes by himself?

How To Solve “Working Together” Problems?

Example: It takes Andy 40 minutes to do a particular job alone. It takes Brenda 50 minutes to do the same job alone. How long would it take them if they worked together?

Word Problem: Work, Rates, Time To Complete A Task

We are given that a person can complete a task alone in 32 hours and with another person they can finish the task in 19 hours. We want to know how long it would take the second person working alone.

Example: Latisha and Ricky work for a computer software company. Together they can write a particular computer program in 19 hours. Latisha van write the program by herself in 32 hours. How long will it take Ricky to write the program alone?

We welcome your feedback, comments and questions about this site or page. Please submit your feedback or enquiries via our Feedback page.

Search form

Derivation of Formulas
Engineering Economy
General Engineering
Spherical Trigonometry
Solid Geometry
Analytic Geometry
Integral Calculus
Differential Equations
Advance Engineering Mathematics
Strength of Materials
Structural Analysis
CE Board: Math
CE Board: Hydro Geo
CE Board: Design
Timber Design
Reinforced Concrete
Geotechnical Engineering
Basic Engineering Math
General Discussions

Breadcrumbs

Work-related problems.

Case 1: Workers have different rates

Work rate = (1 job done) / (Time to finish the job)

Time of doing the job = (1 job done) / (Work rate)

For example Albert can finish a job in A days Bryan can finish the same job in B days Carlo can undo the job in C days

1/A = rate of Albert 1/B = rate of Bryan -1/C = rate of Carlo

Albert and Bryan work together until the job is done: (1/A + 1/B)t = 1 Albert is doing the job while Carlo is undoing it until the job is done: (1/A - 1/C)t = 1

Problem Lejon can finish a job in 6 hours while Romel can do the same job in 3 hours. Working together, how many hours can they finish the job?

Solution Click here to expand or collapse this section Rate of Lejon = 1/6 Rate of Romel = 1/3

$(1/6 + 1/3)t = 1$

$\frac{1}{2}t = 1$

$t = 2 \, \text{ hours}$ answer

Case 2: Workers have equal rates

Work done = no. of workers × time of doing the job

To finish the job

If a job can be done by 10 workers in 5 hours, the work load is 10(5) = 50 man-hours. If 4 workers is doing the job for 6 hours, the work done is 4(6) = 24 man-hours. A remaining of 50 - 24 = 26 man-hours of work still needs to be done.

Problem Eleven men could finish the job in 15 days. Five men started the job and four men were added at the beginning of the sixth day. How many days will it take them to finish the job?

Solution Click here to expand or collapse this section Work load = 11(15) = 165 man-days Work done in 5 days = 5(5) = 25 man-days

Let $x$ = no. of days for them to finish the job $25 + (5 + 4)(x - 5) = 165$

$25 + 9(x - 5) = 165$

$x = 20.56 \, \text{ days}$ answer

Log in or register to post comments
Laws of Exponents and Radicals
Logarithm and Other Important Properties in Algebra
Quadratic Equations in One Variable
Special Products and Factoring
Arithmetic, geometric, and harmonic progressions
Binomial Theorem
System of Equations
Variation / Proportional
Age-related Problems
Clock-related Problems
Digit-Related Problems
Geometry-related Problems
Mixture-related Problems
Money-related Problems
Motion-related Problems
Number-related Problems
A tank is supplied by two pipes A and B and emptied by a third pipe C
Number of Hours for Pipe to Fill the Tank if the Drain is Closed
Probability and Statistics

TIME AND WORK PROBLEMS

1. If a person can do a piece of work in ‘m’ days, he can do ¹⁄ m part of the work in 1 day.

2. If the number of persons engaged to do a piece of work be increased (or decreased) in a certain ratio the time required to do the same work will be decreased (or increased) in the same ratio.

3. If A is twice as good a workman as B, then A will take half the time taken by B to do a certain piece of work.

4. Time and work are always in direct proportion.

more work ----> more time

less work ----> less time

5. A takes m days and B takes n days to complete a work. If they work together, then the formula to find the number of days taken by them to complete the work is

Problem 1 :

A can do a piece of work in 15 days while B can do it in 10 days. How long will they take together to do it?

Using the above formula, if they work together, number of days taken to complete the work is

Problem 2 :

A and B can complete a work in 6 days . B and C can complete the same work in 8 days. C and A can complete in 12 days. How many days will take for A, B and C combined together to complete the same amount of work ?

From the given information, we can have

(A + B)'s 1 day work = ⅙

(B + C)'s 1 day work = ⅛

(A + C)'s 1 day work = ¹⁄₁₂

(A + B + B + C + A + C)'s 1 day work = ⅙ + ⅛ + ¹⁄₁₂

(2A + 2B + 2C)'s 1 day work = ⅙ + ⅛ + ¹⁄₁₂

2 ⋅ (A + B + C)'s 1 day work = ⅙ + ⅛ + ¹⁄₁₂

L.C.M of (6, 8, 12) = 24.

2 (A + B + C)'s 1 day work = ⁴⁄₂₄ + ³⁄₂₄ + ²⁄₂₄

2 (A + B + C)'s 1 day work = ⁹⁄₂₄

2 (A + B + C)'s 1 day work = ⅜

(A + B + C)'s 1 day work = ³⁄₁₆

Time taken by A, B and C together to complete the work is

= 5 ⅓ days

Problem 3 :

A and B can do a work in 15 days, B and C in 30 days and A and C in 18 days. They work together for 9 days and then A left. In how many more days, can B and C finish the remaining work ?

(A + B)'s 1 day work = ¹⁄₁₅

(B + C)'s 1 day work = ¹⁄₃₀

(A + C)'s 1 day work = ¹⁄₁₈

(A + B + B + C + A + C)'s 1 day work = ¹⁄₁₅ + ¹⁄₃₀ + ¹⁄₁₈

(2A + 2B + 2C)'s 1 day work = ¹⁄₁₅ + ¹⁄₃₀ + ¹⁄₁₈

2 (A + B + C)'s 1 day work = ¹⁄₁₅ + ¹⁄₃₀ + ¹⁄₁₈

L.C.M of (15, 30, 18) = 90.

2 (A + B + C)'s 1 day work = ⁶⁄₉₀ + ³⁄₉₀ + ⁵⁄₉₀

2 (A + B + C)'s 1 day work = ¹⁴⁄₉₀

2 (A + B + C)'s 1 day work = ⁷⁄₄₅

(A + B + C)'s 1 day work = ⁷⁄₉₀

Then, the amount of work completed by A, B and C together in 9 days is

= 9 ⋅ ⁷⁄₉₀

= ⁷⁄₁₀

Amount of work left for B and C to complete is

= ³⁄₁₀

Number of days that B will take to finish the work is

= amount of work/part of the work done in 1 day

= ³⁄₁₀ ÷ ¹⁄₃₀

= ³⁄₁₀ ⋅ ³⁰⁄₁

Problem 4 :

A contractor decided to complete the work in 90 days and employed 50 men at the beginning and 20 men additionally after 20 days and got the work completed as per schedule. If he had not employed the additional men, how many extra days would he have needed to complete the work?

The work has to completed in 90 days (as per schedule).

Total no. of men appointed initially = 50.

Given : 50 men have already worked for 20 days and completed a part of the work.

If the remaining work is done by 70 men (50 + 20 = 70), the work can be completed in 70 days and the total work can be completed in 90 days as per the schedule.

Let 'x' be the no. of days required when the remaining work is done by 50 men.

For the remaining work,

70 men ----> 70 days

50 men -----> x days

The above one is a inverse variation.

Because, when no. of men is decreased, no. of days will be increased.

By inverse variation, we have

70 ⋅ 70 = 50 ⋅ x

So, if the remaining work is done by 50 men, it can be completed in 98 days.

So, extra days needed = 98 - 70 = 28 days.

Problem 5 :

Three taps A, B and C can fill a tank in 10, 15 and 20 hours respectively. If A is open all the time and B and C are open for one hour each alternately, find the time taken to fill the tank.

A's 1 hour work = ⅒

B's 1 hour work = ¹⁄₁₅

C's 1 hour work = ¹⁄₂₀

In the first hour, we have

(A + B)'s work = ⅒ + ¹⁄₁₅

(A + B)'s work = ³⁄₃₀ + ²⁄₃₀

(A + B)'s work = ⁵⁄₃₀

(A + B)'s work = ⅙

In the second hour, we have

(A + C)'s work = ⅒ + ¹⁄₂₀

(A + C)'s work = ²⁄₂₀ + ¹⁄₂₀

(A + C)'s work = ³⁄₂₀

Amount of work done in each two hours is

= ¹⁰⁄₆₀ + ⁹⁄₆₀

Amount of work done :

In the first 2 hours : ¹⁹⁄₆₀

In the first 4 hours : ¹⁹⁄₆₀ + ¹⁹⁄₆₀ = ³⁸⁄₆₀

In the first 6 hours : ¹⁹⁄₆₀ + ¹⁹⁄₆₀ + ¹⁹⁄₆₀ = ⁵⁷⁄₆₀

After 6 hours, the remaining work will be

= ¹⁄₂₀

¹⁄₂₀ is the small amount of work left and A alone can complete this.

Time taken by A to complete this 1/20 part of the work is

= amount of work/part of work done in 1 hour

= ¹⁄₂₀ ÷ ⅒

= ¹⁄₂₀ ⋅ ¹⁰⁄₁

= ½ hours

So, A will will take half an hour (or 30 minutes) to complete the remaining work ¹⁄₂₀ .

So, total time taken to complete the work is

= 6 hours + 30 minutes

= 6 ½ hours

Kindly mail your feedback to [email protected]

We always appreciate your feedback.

Sat Math Practice
SAT Math Worksheets
PEMDAS Rule
BODMAS rule
GEMDAS Order of Operations
Math Calculators
Transformations of Functions
Order of rotational symmetry
Lines of symmetry
Compound Angles
Quantitative Aptitude Tricks
Trigonometric ratio table
Word Problems
Times Table Shortcuts
10th CBSE solution
PSAT Math Preparation
Privacy Policy
Laws of Exponents

Recent Articles

Digital sat math problems and solutions (part - 21).

Sep 05, 24 11:17 AM

Digital SAT Math Problems and Solutions (Part - 19)

Sep 04, 24 10:53 PM

Digital SAT Math Problems and Solutions (Part - 18)

Sep 04, 24 10:49 PM

Wilmette makes headway on rat problem, but there’s more work to do

“The response has been really strong, which is very important to getting this problem under control,” he said.

Already a subscriber? You can make a tax-deductible donation at any time.

Joe Coughlin

Joe Coughlin is a co-founder and the editor in chief of The Record. He leads investigative reporting and reports on anything else needed. Joe has been recognized for his investigative reporting and sports reporting, feature writing and photojournalism. Follow Joe on Twitter @joec2319

The Record is a nonprofit, community news site from former editors of 22nd Century Media

New trier high school goes into 'brief' lockdown for undefined 'emergency', athlete of the week: 10 questions with kit kat mcgregor, new trier volleyball, artist ready to begin wilmette mural, which should be finished this month.

The Record: Shortcut

Information

We use cookies to collect anonymous data to help us improve your site browsing experience.

Click 'Accept all cookies' to agree to all cookies that collect anonymous data. To only allow the cookies that make the site work, click 'Use essential cookies only.' Visit 'Set cookie preferences' to control specific cookies.

Your cookie preferences have been saved. You can change your cookie settings at any time.

Programme for Government - First Minister's speech - 4 September 2024

Presiding Officer,

This year, Parliament marks the 25 th anniversary of its opening, and I have witnessed every previous Programme for Government being announced – albeit from different places across the Parliamentary Chamber.

Today, however, is the first time that I present a Programme for Government. It is an extraordinary privilege to do so and to have the opportunity to further shape the direction of our country.

I do so in a spirit that recognises that we all come from different political traditions. I believe that Scotland would best be able to progress, as an independent country, where the issues we address in this programme can be more effectively resolved. Others take the opposite view. We have a range of priorities and perspectives, but fundamentally we are all here to contribute to creating the best future we can for Scotland.

My Government does not command a majority in this Parliament. We have to work with others to make progress on our agenda. I therefore set out this Programme for Government with a commitment to work across this Chamber to seek common ground with others.

I extend the invitation to colleagues to work together to find that common ground.

A quarter of a century after its creation, this Parliament faces some of its toughest tests.

We are all aware of the problems and difficulties that have been caused by 14 years of austerity driven by the UK Government.

We are all aware of the acute challenges that are faced due to the impact of sky-high inflation and the failure of the UK Government to adequately increase Budgets to address that fact.

We know that, because we all can see the pressure on our public services and also because the new Chancellor of the Exchequer has made that very point clear to the House of Commons.

The Scottish Government has set out to Parliament the difficult decisions we have to take to address those circumstances.

Today, I set out how, within that challenging situation, my Government will deliver for the people of Scotland.

This Programme for Government sets out – simply and clearly – our intentions for the next 12 months.

Its purpose is to ensure this government spends every day delivering for the people of Scotland.

The commitments in it are practical, not partisan. They are affordable, impactful and deliverable. Together, they reflect my optimism, that even though we face an incredibly challenging set of circumstances at this moment, the inherent strengths of Scotland, our people and our communities, can create great possibilities for our country.

When I became the First Minister, I made clear that my Government would focus on four priorities:

eradicating child poverty,
building prosperity,
improving our public services,
and protecting the planet.

Child poverty is first and foremost in these priorities.

No child should have their opportunities, their development, their health and wellbeing, and their future curtailed by the material wealth of their family. Not ever, and certainly not today, in a modern, prosperous society like Scotland.

This is not only the moral compass of my Government, it is the greatest investment in our country’s future that we can possibly make.

It is the route to enabling greater participation in our economy and our society. It is the route to enabling more people to fulfil their potential and to be contributors to our country.

We have dedicated roughly £3 billion a year to eradicating poverty and mitigating the impacts of the cost-of-living crisis.

We have established and increased our widely praised Scottish Child Payment, expanded funded early learning and childcare, and committed around £1.2 billion to mitigate the impacts of 14 years of UK welfare policy.

These measures, which are key to increasing family incomes and enabling those greater levels of participation in our economy, are central to our Programme for Government.

The Child Poverty Action Group estimates that low-income families in Scotland will be around £28,000 better off by the time their child turns 18, compared to families across the UK.

And analysis by the Scottish Government estimates that around 100,000 children will be kept out of relative poverty this year.

These achievements are significant and show the difference that we are making. But our goal is not just to keep some children out of poverty, or only to make child poverty less acute.

Our goal is to lift every child in Scotland who is in poverty out of it.

So we must do more.

We know that we cannot address child poverty without addressing family poverty.

We know that families thrive when they are supported by coordinated, holistic services that meet their needs and are easy to access.

Many amazing and dedicated practitioners are already working tirelessly to connect services and adapt them to the needs of the families that they support.

We must create a system of whole-family support that is available across the country. We must ensure this system is easy to access, well-connected and responsive to families’ needs.

Over the coming year, we will work with partners to enable greater local flexibility, so that services can be more easily tailored to the needs of the families that they support.

We will look at what budgets can be pooled and what reporting can be streamlined. This will involve working closely with our local authority partners, other public services and the Third Sector, to align services and ensure there is a focus across our public services on meeting the needs and supporting the resilience of families.

We will consider where greater investment is needed.

And we will use the learning of what is already working from our pilot areas.

Some of that evidence comes from, for example, the Early Adopter Community project in Dundee.

In that project, keyworkers have been engaging with members of the public who face obstacles to entering the labour market.

By providing focused childcare support, advice on eligibility for benefit provision, and employability support, individuals are being supported into the labour market.

These individuals are sustaining employment and experiencing a number of benefits to their financial and mental wellbeing.

The key objective of the approach that we will take will be to deliver significant reform of the work of public services to deliver whole-family support extensively across the country.

This will create the conditions that support more parents into employment and reinforce our work to eradicate child poverty.

Key to the work on whole family support will be a focus on prevention and early intervention, those small supports early on that can pay big dividends down the line.

This includes pregnancy and the first years of a child’s life.

Addressing risks and problems at this stage can have positive impacts that last through to adulthood.

It can support healthy development, prevent illness and ease future pressure on services, making the entire system more sustainable.

So, in the coming year, we will ensure that more women get to know, and receive care and support from, the same midwifery team from pregnancy through birth.

We will invest nearly £1 billion a year in affordable, high-quality and funded early learning and childcare.

And we commit to supporting early development and reducing developmental concerns at 27-30 months by a quarter by 2030.

We will support schools to reduce the poverty-related attainment gap across every local authority each year between now and 2026.

And we will ensure that, when young people are ready to enter the workplace, they have the learning, the skills and the opportunities to succeed.

We will invest in community-based youth work and improve careers support so that there is better information on career choices.

For those households struggling now, boosting financial security and cutting costs is one of the most direct things that we can do to support them out of poverty.

So, we will expand advice in accessible settings, including community centres and hospitals. This will expand a programme that, in its first year, helped over 5,500 people access financial gains of over £7.5 million to support their families.

We will also complete the national rollout of our Carer Support Payment, which will support over 100,000 carers this financial year – including, for the first-time ever, some in full-time education.

The effect of this provision will be to enable much greater participation in education for those with caring responsibilities, which will greatly increase the opportunities for those individuals to make an economic contribution to our society.

That effort to stimulate greater economic participation lies at the heart of measures in this Programme for Government.

A key aspect also of our Programme to support families is to ensure that we take effective action to enable people to have a safe and secure place to call their home.

The tragic Grenfell Tower fire emphasised how important building and fire safety is. Keeping residents and homeowners safe is our priority and we are taking action to protect lives by ensuring the assessment and remediation of buildings with potentially unsafe cladding is carried out.

We will also carefully consider all the recommendations in the Grenfell Tower Inquiry’s report

This year we will invest nearly £600 million in affordable housing, including an additional £40 million to bring existing homes into affordable use.

There will be a strong focus on working with partners to enable existing accommodation that is not currently in use to be made available as swiftly as possible to meet the need for housing – ensuring we take every step we can to boost the availability of housing as quickly as we possibly can.

We will provide a further £100 million to support the construction of around 2,800 mid-market rent homes.

And in progressing our proposals for rent controls, we will introduce amendments at stage 2 of the Housing Bill to ensure that tenants have the protection they need and that Scotland is able to attract more investment to supplement the investment we are making through the public finances.

These commitments are central to our efforts to tackle poverty, but they are also inextricably linked to our efforts to increase national prosperity with a strong, green, wellbeing economy.

It takes thriving businesses, large and small, to sustain our families and our communities.

We are already making significant progress in this area. Since 2007, GDP per person in Scotland has grown by 11%, compared to the UK’s 6%.

Productivity has grown at an average rate more than double that of the UK.

And, last year, earnings in Scotland grew more quickly than in any other part of the UK. This includes London and the South East.

But again, we must go further.

It is not enough to simply have a strong economy. True prosperity goes beyond pounds and pence.

It means an economy that is inclusive, that supports people into work, attracts investment, promotes entrepreneurs and innovators, and furthers our work on our path to net zero.

Key to this will be increasing the levels of infrastructure investment and creating the right conditions for business investment.

For the past 9 years, we have been the UK’s top destination outside of London for foreign direct investment.

This Government will be focused on delivering investment-friendly policies and support, such as the Scottish National Investment Bank, which will help to build on the strong performance we have built to date.

Last year, the bank supported 1,850 jobs by investing in companies with over £92 million of supply chain spend in Scotland.

Its £60 million investment in Thriving Investments Mid Market Rent fund will help deliver affordable, quality rental homes.

Put simply, this means tenants benefitting from the scheme will pay lower rents but the economy will be boosted by the development activity involved.

To ensure Scotland remains a premier location for investment, we will align government and public bodies behind a co-ordinated programme to attract investment in priority areas, such as net zero, housing and infrastructure.

We will build on recent successes, such as the Sumitomo and Ardersier projects, to promote our national project pipeline of investment opportunities.

We will develop two Green Freeports and establish two new Investment Zones.

And – with our commitment of up to £500 million of investment – we will seek to generate at least £1.5 billion in private investment to support the offshore wind sector.

A critical element of ensuring that Scotland is attractive to investors, means that we must intensify our support to Scotland’s innovators and entrepreneurs as part of our work to become a start-up and scale-up nation.

So this year, we will maximise the impact of our national network of startup support, our Techscaler programme.

The programme has already supported start-ups to raise £70 million in investment.

We will also work with organisations like Scottish Enterprise, the National Manufacturing Institute for Scotland and the National Robotarium to create new opportunities for our most promising 'deep tech' companies.

We will ensure our universities can contribute to international-leading research and economic growth and support the development of business clusters in areas such as digital and AI, life sciences and the energy transition.

Small and medium-sized businesses are the backbone of our economy, so we will deliver the commitments set out in the New Deal for Business.

We will empower decision-making through Regional Economic Partnerships and sign the Falkirk and Grangemouth and Argyll & Bute Regional Growth Deals.

We will tackle economic inactivity and skills shortages in our workforce and remove barriers to employment.

This includes leading a new, national approach to skills planning and introducing the Post-School Education Reform Bill to simplify the post-school funding body landscape.

Specialist support for disabled people will be enhanced across all local authorities by the summer of 2025.

We will expand Scotland’s Migration Service and continue to make the case for tailored migration routes – including a Rural Visa Pilot to support rural employers to recruit the people they need.

We will support Scotland’s culture sector and creative industries, which are key to our economy, our culture and national identity, and we recognise the need for the artistic and creative community to be well supported for the future.

A review of Creative Scotland will be undertaken to ensure the appropriate approach is in place to meet the needs of the sector and I am pleased to confirm to Parliament that the resources required to enable Creative Scotland to continue the work of the Open Fund are now available for them to distribute.

And we will continue to invest in our national infrastructure – the transport and digital networks that enable our economy to thrive.

This year, over 20,000 premises will be connected to gigabit capable broadband across Scotland in areas of market failure.

We will progress dualling the A9, with construction expected to start before the end of this year on the Tomatin to Moy stretch and the procurement process already underway on the Tay Crossing to Ballinluig stretch.

We will deliver 3 of the 6 new major ferries currently under construction at the present time, and we will progress the procurement of 7 new electric ferries as part of the Small Vessel Programme.

We will also continue to invest in our rail network and upgrade and reconfigure power supplies to support further electrification of our railways.

All of this will improve access to and from our rural and island communities, improving transportation safety, journey times and reliability, and generate economic growth.

It will also enable delivery of our valuable public services and ensure people in every corner of Scotland have access to the high quality services they need, when they need them.

Public services touch every aspect of our day-to-day lives. They support our families. They enable our economy to grow and to thrive.

Key to public services is ensuring everyone has access to high-quality services that are right for them. Those services must be easy to access and to navigate.

Nowhere is this more notable than with our National Health Service.

We have seen terrific successes in our NHS, including the best performing core A&E departments anywhere in the UK.

We need to ensure our NHS has the resources it needs, both for today and for the years to come.

We will increase the Boards’ baseline funding to reduce waiting list backlogs.

We will deliver around 20,000 more orthopaedic, ophthalmology and general surgery procedures annually in our new National Treatment Centres.

And free up 210,000 planned care outpatient appointments through our Centre for Sustainable Delivery programmes – thus eliminating unnecessary hospital attendances.

We will also reform primary care – increasing capacity and access to general practice, community pharmacy, dental and community eyecare services by the end of 2026.

And, backed by £120 million of additional funding for NHS Boards, we will support continued improvements across a range of mental health services and treatments.

This includes meeting the Child and Adolescent Mental Health Services waiting times standard nationally and clearing backlogs by December 2025.

We will intensify our work to tackle delayed discharge.

No one should remain in hospital any longer than they need to. So we will standardise best practice and an integrated approach, from the time a person enters hospital through to their timely discharge.

This will ensure everyone can recover in the best, least intensive setting for them, whilst also making room in hospital for those who need it.

But in this challenging fiscal environment, we cannot deliver public services as we did in years past.

We must change the model of service delivery to promote positive outcomes, prioritise prevention and reduce demand for future services.

Once again, intelligent investment and innovation will be key here.

For example, the Scottish Government invested £4 million to pilot Rapid Cancer Diagnostic Services across 5 NHS boards.

This service works through primary care to provide a quicker diagnosis to people experiencing non-specific symptoms.

Impressively, an evaluation of the pilot in NHS Fife and in NHS Dumfries and Galloway found the estimated time to diagnosis was roughly 65 days faster than via the more usual route of a general surgery clinic.

Earlier diagnosis means better outcomes, less intensive treatment and less strain on the system. It means Scotland, as a whole, is healthier.

We have similar ambitions for all of our public services as part of our 10-year Public Service Reform programme, which will guide our approach.

And every area of this government is committed to delivering reform consistent with those principles.

As part of this ambitious programme, we will work with local authorities to boost school standards – with a focus on attendance, behaviour, and the curriculum.

We will implement the Curriculum Improvement Cycle and progress with qualifications reform.

And we will reform our national education bodies to drive improvement, raise standards and ensure that the needs of learners are always at the forefront of our work.

We attach the greatest significance to the safety of our communities so we will work with Police Scotland to ensure that that remains the case.

And because our public services are only as strong as the people who deliver them, we will continue to award fair pay settlements, reduce workloads and improve conditions for our public service employees.

We will review and reform the Junior Doctor and Dentist contract, progress towards the 36-hour working week for Agenda for Change staff, and provide local authorities with £145 million to protect teacher numbers.

But every one of these important actions – and indeed everything I have mentioned so far – will be rendered ineffective if we do not also address the greatest existential threat of our times.

We must take effective action to tackle the twin crises of climate change and biodiversity loss.

It is absolutely essential that we protect our planet by reducing emissions, restoring our natural environment and investing in adaptations that will protect us from the impacts of climate change.

While we are decarbonising at a faster rate than the rest of the UK, the most difficult part of the journey lies ahead.

The world’s global temperature has now pushed past the internationally agreed 1.5 degrees Celsius.

Ten of our hottest years have come in the last 20, and the increased frequency of storms and floods is already having a real impact on communities and key sectors.

In addition to this, the 2023 State of Nature in Scotland report found that monitored species have declined by 15% over the last 30 years.

We are already making real progress protecting our environment and helping it to recover. 75% of all new UK woodlands are here in Scotland.

But we must adapt to the changes in our environment. That is why we will take forward our National Adaptation Plan.

We will work with Scottish Water to improve the resilience of our water and sewerage systems to intense rainfall and drought.

And we will restore at least 10,000 hectares of degraded peatland and create at least 10,000 hectares of woodlands.

We will also bring forward a Natural Environment Bill to support delivery of our net zero and biodiversity goals.

Climate change legislation that will enable 5-year carbon budgets to be set and delivered will be introduced.

This, along with our Climate Change Plan and sectoral Just Transition Plans, will chart our course to net zero by 2045.

Tackling the climate emergency is not only a danger that must be recognised and managed. It is an imperative that should motivate us to change.

Scotland is a land of remarkable innovation and abundant natural resources.

We can tackle climate change whilst at the same time growing our economy. Indeed, we have been doing it for decades.

Between 1990 and 2022, Scotland's economy grew by 67% in real terms. In that same time, we cut our greenhouse gas emissions in half.

And last year, the Scottish National Investment Bank – in the course of investing in businesses and projects that support our economy – also avoided, reduced or removed over 52,000 tonnes of carbon dioxide equivalent across its portfolio.

In 2022, renewable technologies generated the equivalent of 113% of Scotland’s overall electricity consumption.

In 2024, our capacity for renewable energy generation increased to 15.4 gigawatts, and with the projects currently in the planning pipeline, we have the potential to produce more than three times that.

We will shortly publish our Energy Strategy and Just Transition Plan. By delivering the commitments within it we will again double our ambitions for renewable energy generation.

As part of this, we are acting to speed up the planning and consenting regime for renewable energy generation to provide certainty to the market and to stimulate private investment.

We will invest £9 million to support Scotland’s manufacturing industries to invest in energy efficiency and decarbonisation projects.

And we will work with the UK Government to deliver the infrastructure required for a net zero emissions energy system, including by providing £2 million – as part of the promised support for carbon capture of up to £80 million – to support the Acorn Carbon transport and storage project and securing a positive future for Grangemouth.

We need to take forward careful stewardship of our oil and gas sector to ensure a sector that contributes significantly to the economic health of Scotland at this moment is able to make the transition effectively to net zero.

The expertise of the sector will be vital both to the future of the industry and to our transition to net zero.

We need to keep those skills in Scotland as we move toward a green economy. For we are in the midst of a renewables revolution here in Scotland.

Alongside our investment in renewables, we will support households and communities to reduce emissions.

More than a third of the population already benefit from our offer of free bus travel.

We have some of the most generous grants and loans in the UK to support the move to clean heating. And we are within reach of 6,000 electric vehicle charging network points in 2024, two years ahead of schedule.

This year, we will conclude the review of our New Build Heat Standard and bring forward a Heat in Buildings Bill.

The Bill will set a long-term direction of travel that is deliverable and affordable for households and businesses. It will also provide certainty to building owners and the supply chain.

Through the work of Home Energy Scotland and the support available through our Warmer Homes Scotland scheme, we will take forward measures to ensure we offer practical solutions to encourage energy efficiency and to enable families to stay warm.

The purpose of the Heat in Buildings Bill must be to enable practical assistance to be made available to households and businesses to support energy efficiency and to improve the quality of heating systems.

We will also set a clear timetable for the delivery of roughly 24,000 additional electric vehicle charge points by 2030.

And we will make it easier for people to walk, wheel or cycle through our Active Travel Infrastructure Fund, the National Cycle Network and our People and Place Programme.

These commitments are good for our communities and good for our environment. Plain and simple, they are good for Scotland.

Beyond the question of my Government’s priorities and the specifics of the programme, there is one further question I want to address.

My Ministers are in the public service.

I want my Government to set the highest standard of propriety and integrity.

I want trust to be at the heart of our relationship with the people of Scotland.

That is why I intend to make changes to strengthen the Scottish Ministerial Code.

Investigations into alleged breaches of the Code will no longer happen only at the instruction of the First Minster.

Independent Advisers will be able to launch their own investigations whenever they feel it is warranted.

And where there has been a breach, they will be able to advise on appropriate sanctions.

These changes will significantly strengthen the role of the independent advisers, whose Terms of Reference will also be published.

I expect to publish the new code by the end of this year.

Presiding Officer, Scotland is a country of many strengths.

Our economy is founded on industries of global reach in energy, financial services, food and drink, tourism, life sciences and advanced manufacturing.

Our education system is high performing and includes a number of world class universities.

Our natural environment is of the highest quality and provides the basis for so many of our economic strengths.

Our talented and creative population is our greatest asset, enhanced by those who choose to make their future here.

Our society is bound together by a strong sense of social justice, of acting together to build the common good to ensure everyone in our country is able to fulfil their potential.

Yes, we face challenges. But if properly focused and motivated, I am optimistic and confident that the inherent strengths of our country will help us overcome those challenges.

With good will, and with a relentless focus on delivering for the people of Scotland, I believe the resources available to us can be used to help us eradicate child poverty, build prosperity, improve our public services and play our part in protecting the planet.

That is the focus of this Programme for Government, and I commend it to Parliament.

There is a problem

Thanks for your feedback

Your feedback helps us to improve this website. Do not give any personal information because we cannot reply to you directly.

IMAGES

Effective problem-solving strategies
3 reasons why problem-solving is important in the workplace
How to Tell if You are the Source of Your Problems at Work
Math Work Problems (video lessons, examples and solutions)
Problem Solving is a Must Have In The Workplace, Here is Why
The Problem Determines the Solution

VIDEO

This Technique can solve your problem! #shaleenshrotriya #businesscoach #solution #coaching
5 Ways to Level Up Your Productivity
Work Energy Problem
4.6 Problem 4 based on Reciprocating air compressors work done
Work Rate Word Problem
Stop Making A Problem Which Doesnt Exist!

COMMENTS

Work done by force
B ased on the above formula, work done by force and there is a displacement. The correct answer is C. 1 3. Andrew pushes an object with force of 20 N so the object moves in circular motion with a radius of 7 meters. Determine the work done by Andrew for two times circular motion. A. 0 Joule
Work Done by a Force (examples, solutions, videos, notes)
Work Done By a Constant Force & Energy Transfer. Example 1: A man applies a force of 700 N to a crate and pushes it through a distance of 200 cm. Calculate the amount of work done by the man. Example 2: Another man pushes a crate as shown with a force of 550 N and soes 2.2 kJ of work.
9.10 Rate Word Problems: Work and Time
The equation used to solve problems of this type is one of reciprocals. It is derived as follows: rate ×time = work done rate × time = work done. For this problem: Felicia's rate: F rate × 4 h = 1 room Katy's rate: Krate × 12 h = 1 room Isolating for their rates: F = 1 4 h and K = 1 12 h Felicia's rate: F rate × 4 h = 1 room Katy's rate: K ...
"Work" Word Problems
As you can see in the above example, "work" problems commonly create rational equations. But the equations themselves are usually pretty simple to solve. One pipe can fill a pool 1.25 times as fast as a second pipe. When both pipes are opened, they fill the pool in five hours.
Algebra Work Problems (solutions, examples, videos)
Solution: Step 1: Assign variables: Let x = time taken to fill up the tank. Step 2: Use the formula: Since pipe C drains the water it is subtracted. Step 3: Solve the equation. The LCM of 3, 4 and 5 is 60. Multiply both sides with 60. Answer: The time taken to fill the tank is hours.
7.1 Work
We can solve this problem by substituting the given values into the definition of work done on an object by a constant force, stated in the equation W = F d cos θ W = F d cos θ. The force, angle, and displacement are given, so that only the work W is unknown. Solution The equation for the work is
Calculating the Amount of Work Done by Forces
Calculating the Amount of Work Done by Forces. In a previous part of Lesson 1, work was described as taking place when a force acts upon an object to cause a displacement. When a force acts to cause an object to be displaced, three quantities must be known in order to calculate the work.
Work Rate Problems with Solutions
The work done by John alone is given by. t × (1 / 1.5) The work done by Linda alone is given by t × (1 / 2) When the two work together, their work will be added. Hence ... 1 / 20) = 1 Solve for t t = 3.6 hours. Problem 4: A swimming pool can be filled by pipe A in 3 hours and by pipe B in 6 hours, each pump working on its own. At 9 am pump A ...
Work Word Problems (video lessons, examples, solutions)
Solution: Step 1: Assign variables: Let x = time taken by Peter. Step 2: Use the formula: Step 3: Solve the equation. Multiply both sides with 30 x. Answer: The time taken for Peter to paint the fence alone is hours. Example 2: Jim can dig a hole by himself in 12 hours.
How to Master Work Problems: A Comprehensive Step-by-Step Guide
Understanding work problems in mathematics often involves dealing with scenarios where different people (or machines) contribute to completing a task. These problems can be solved by using the formula $W=R×T$, where $W$ is Work, $R$ is Rate, and $T$ is Time. Here is a step-by-step guide to help you understand and solve these problems:
Physics Work Problems for High Schools
Problem (11): A $5-{\rm kg}$ box, initially at rest, slides $2.5\,{\rm m}$ down a ramp of angle $30^\circ$. The coefficient of friction between the box and the incline is $\mu_k=0.435$. Determine (a) the work done by the gravity force, (b) the work done by the frictional force, and (c) the work done by the normal force exerted by the surface.
Work, Energy, and Power Problem Sets
Problem 2: Hans Full is pulling on a rope to drag his backpack to school across the ice. He pulls upwards and rightwards with a force of 22.9 Newtons at an angle of 35 degrees above the horizontal to drag his backpack a horizontal distance of 129 meters to the right. Determine the work (in Joules) done upon the backpack.
40 problem-solving techniques and processes
7. Solution evaluation. 1. Problem identification. The first stage of any problem solving process is to identify the problem (s) you need to solve. This often looks like using group discussions and activities to help a group surface and effectively articulate the challenges they're facing and wish to resolve.
Time and Work Problems
This math video tutorial focuses on solving work and time problems using simple tricks and shortcuts. It contains a simple formula that you can use with the...
Word Problems on Time and Work
Then, work done by A in 1 day = 1/nᵗʰ part of the work. (ii) Suppose that the work done by A in 1 day is $\frac{1}{n}$ Then, time taken by A to finish the whole work = n days. Problems on Time and Work : 1. Aaron alone can finish a piece of work in 12 days and Brandon alone can do it in 15 days. If both of them work at it together, how ...
Work Done Problems
In this article, we will learn how we can solve the work done problems. Work Done Related Important Formula. In time and work, we will learn to calculate and determine the number of hours needed to complete a task as well as the amount of work completed in a specific period of time. We are aware that a person's productivity is closely ...
Work Problems
The next step to solving this problem demands that we find values for the last column, "Work." Imagine if Ben worked for 2 hours. We would multiply 1 ⁄ 4 times 2 and get 1 ⁄ 2 , which would mean only 1 ⁄ 2 of the job was done.
Time and Work Formula and Solved Problems
FORMULAS. The basic formula for solving is: 1/r + 1/s = 1/h. Let us take a case, say a person Hrithik. Let us say that in 1 day Hrithik will do 1/20 th of the work and 1 day Dhoni will do 1/30 th of the work. Now if they are working together they will be doing 1/20 + 1/30 = 5/60 = 1/12 th of the work in 1 day. Now try to analyze, if two persons ...
Math Work Problems (video lessons, examples and solutions)
Solution: Step 1: Assign variables: Let x = time to mow lawn together. Step 2: Use the formula: Step 3: Solve the equation. The LCM of 40 and 60 is 120. Multiply both sides with 120. Answer: The time taken for both of them to mow the lawn together is 24 minutes.
How To Put Problem-Solving Skills To Work in 6 Steps
Here are the basic steps involved in problem-solving: 1. Define the problem. The first step is to analyze the situation carefully to learn more about the problem. A single situation may solve multiple problems. Identify each problem and determine its cause. Try to anticipate the behavior and response of those affected by the problem.
Work-related Problems
Work-related Problems. Case 1: Workers have different rates. Work rate × Time to finish the job = 1 job done. Work rate = (1 job done) / (Time to finish the job) Time of doing the job = (1 job done) / (Work rate) For example. Albert can finish a job in A days. Bryan can finish the same job in B days. Carlo can undo the job in C days.
TIME AND WORK PROBLEMS
TIME AND WORK PROBLEMS. 1. If a person can do a piece of work in 'm' days, he can do ¹⁄m part of the work in 1 day. 2. If the number of persons engaged to do a piece of work be increased (or decreased) in a certain ratio the time required to do the same work will be decreased (or increased) in the same ratio. 3.
Wilmette makes headway on rat problem, but there's more work to do
Officials blame multiple years of mild winters for increased rat activity in Wilmette. Numerous reports of infestations led the Village of Wilmette to urge residents for help combatting what Braiman called the town's worst rat problem in recent memory.. To enable that help, the Village offered resources, such as free property inspections and guidance and up to 100% financial reimbursement to ...
Could these 'motionless' turbines solve wind energy's noise and
"Our "motionless" wind energy technology is designed to work seamlessly alongside solar systems, maximising the renewable energy output from rooftops while helping address challenges like ...
Workshop to focus on problem solving using the 'Plan, Do, Check, Act
The workshop will introduce attendees to the "Plan, Do, Check, Act" Kata problem-solving process, which will provide participants the tools for streamlining their shops and codes. "It's a class that teaches employees how to use a system that helps give them some control of their work space," said Paul Sherman, lead instructor, Code 100TO ...
First Minister's speech
Key to the work on whole family support will be a focus on prevention and early intervention, those small supports early on that can pay big dividends down the line. This includes pregnancy and the first years of a child's life. Addressing risks and problems at this stage can have positive impacts that last through to adulthood.