example of kinetic energy problem solving

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Practice Problems on Kinetic Energy

When work is done by a force on an object. It acquires energy, it can be any form. Energy can take on many forms and can be converted from one form to another form. Potential energy, electric potential energy, kinetic energy, etc. are some examples of different types of energy. Kinetic energy comes when the object starts moving. This energy is due to motion. Although this energy is due to motion, this energy is not created. It is usually converted from one type of energy to another type. Let’s look at this concept in detail. 

Kinetic Energy

If an object is stationary, and we want to put that object into motion. We need to apply force. Any type of acceleration requires some force. When this force is applied, work is done on the object. When the work is done on an object, this means energy is getting transferred to the object is one form or another. Force can be removed once the object is in motion, but till the time force was applied on the object. The work that was done during that time is converted into energy. 

Kinetic energy is the energy an object acquires by virtue of its motion. 

This energy can be transferred from one object to another. For example, a moving ball hitting a stationary ball might cause the other ball to move. In this situation, some kinetic energy of the ball is transferred to another ball. 

Formula of Kinetic Energy 

To calculate the kinetic energy of the object, let’s consider a scenario where a force F, is acting on an object of mass M. In this case, the object starts moving with the acceleration “a” and covers a distance of “d”. 

Work done in this case will be, 

The acceleration “a” can be replaced using an equation of motion. 

v 2 = u 2 + 2a.d

⇒v 2 – u 2 = 2a.d

⇒ [Tex]\frac{v^2 – u^2}{2a}[/Tex]  = d

Substituting the value of “d” in the equation, 

⇒ W =  [Tex]m.d.\frac{v^2 – u^2}{2d}[/Tex]

⇒W =  [Tex]m.\frac{v^2 – u^2}{2}[/Tex]

⇒W =  [Tex]\frac{1}{2}m(v^2 – u^2)[/Tex]

So, this whole work done is converted into the K.E of the object. 

In case, initial velocity u = 0, 

K.E =  [Tex]\frac{1}{2}mv^2[/Tex]

One can also say, the network work done on the system is equal to the change in kinetic energy of the object. 

Note:  1. Kinetic energy depends on the velocity of the object squared. This means, when th velocity of the object is doubled, its kinetic energy becomes four times.  2. K.E must always have zero or positive values.  3. Kinetic energy is a scalar quantity, and it is expressed in Joules. 

Sample Problems

Question 1: A ball has a mass of 2Kg, suppose it travels at 10m/s. Find the kinetic energy possessed by it. 

Answer: 

Given: m = 2Kg, and v = 10m/s  The KE is given by,  K.E =  [Tex]\frac{1}{2}mv^2[/Tex] K.E =  [Tex]\frac{1}{2}mv^2[/Tex] ⇒ K.E =  [Tex]\frac{1}{2}(2)(10)^2[/Tex] ⇒ K.E = 100J

Question 2: A ball has a mass of 10Kg, suppose it travels at 100m/s. Find the kinetic energy possessed by it. 

Given: m = 10Kg, and v = 100m/s  The KE is given by,  K.E =  [Tex]\frac{1}{2}mv^2[/Tex] K.E =  [Tex]\frac{1}{2}mv^2[/Tex] ⇒ K.E =  [Tex]\frac{1}{2}(10)(100)^2[/Tex] ⇒ K.E = 50000J

Question 3: A spaceship has a mass of 20000Kg, suppose it travels at 10m/s. Find the kinetic energy possessed by it. 

Given: m = 20000Kg, and v = 10m/s  The KE is given by,  K.E =  [Tex]\frac{1}{2}mv^2[/Tex] K.E =  [Tex]\frac{1}{2}mv^2[/Tex] ⇒ K.E =  [Tex]\frac{1}{2}(20000)(10)^2[/Tex] ⇒ K.E = 10 6 J

Question 4: Work done by a force on a moving object is 100J. It was traveling at a speed of 2 m/s. Find the new speed of the object if the mass of the object is 2Kg.

Answer:  Given: W = 100J Work done by the force is equal to the change in kinetic energy.  W =  [Tex]\frac{1}{2}m(v^2 – u^2)[/Tex] Given, u = 2 m/s and v = ?, m = 2kg.  Plugging the values in the given equation,  W =  [Tex]\frac{1}{2}m(v^2 – u^2)[/Tex] ⇒  [Tex]100 = \frac{1}{2}(2)(v^2 – 2^2)[/Tex] ⇒ [Tex]100 = v^2 – 2^2 \\ = 104 = v^2 \\ = v = \sqrt{104}  \text{ m/s}[/Tex] [Tex]v = \sqrt{104} \text{ m/s}[/Tex]

Question 5: Work done by a force on a moving object is -50J. It was traveling at a speed of 10m/s. Find the new speed of the object if the mass of the object is 2Kg.

Given: W = -50J Work done by the force is equal to the change in kinetic energy.  W =  [Tex]\frac{1}{2}m(v^2 – u^2)[/Tex] Given, u = 10m/s and v = ? . m = 2kg.  Plugging the values in the given equation,  W =  [Tex]\frac{1}{2}m(v^2 – u^2)[/Tex] ⇒  [Tex]-50 = \frac{1}{2}(2)(v^2 – 10^2)[/Tex] ⇒  [Tex]-50 = v^2 – 10^2 \\ = 50 = v^2 \\ = v = \sqrt{50}  \\ = v = 5\sqrt{2} \text{ m/s}[/Tex] The speed is decreased because the work done was negative. This means that the force was acting opposite to the block and velocity was decreased. 

Question 6: Suppose a 1000Kg was traveling at a speed of 10m/s. Now, this mass transfers all its energy to a mass of 10Kg. What will be the velocity of the 10Kg mass after being hit by it? 

KE is given by the formula, K.E =  [Tex]\frac{1}{2}mv^2     [/Tex]   KE of the heavier object  M = 1000Kg and v = 10m/s  K.E =  [Tex]\frac{1}{2}mv^2     [/Tex]   ⇒ K.E =  [Tex]\frac{1}{2}(1000)(10)^2[/Tex] ⇒K.E = 50,000J  Now this energy is transferred to another ball.  m = 10Kg and v = ? 50,000 =  [Tex]\frac{1}{2}(10)v^2[/Tex] ⇒ 10,000 = v 2 ⇒ v = 100 m/s

Question 7: Suppose a 10Kg mass was traveling at a speed of 100m/s. Now, this mass transfers all its energy to a mass of 20Kg. What will be the velocity of the 20Kg mass after being hit by it? 

To find the velocity of the 20 kg mass after being hit by the 10 kg mass, we use the principle of conservation of momentum. According to the conservation of momentum, the total momentum before the collision is equal to the total momentum after the collision, assuming no external forces are acting on the system. m 1 = 10kg v 1 = 100m/s m 2 =20kg v 2 = ? The total momentum before the collision (P initial) is : = m 1 x v 1 = 10 x 100 = 1000 Kgm/s According to the conservation of momentum: P initial = P final m 1 x v 1 = m 2 x v 2 10 x 100 = 20 x v 2 1000 = 20 x v 2 v 2 = 1000/20 = 50m/s So, the velocity of the 20 kg mass after being hit by the 10 kg mass is 50m/s.

Question 8: Suppose a 10Kg block was kept at 20m height. Now, this block is dropped. Find out the velocity of the block just before it hits the ground.

The block of 10Kg is kept at a height of 20m.  The potential energy of the block will be,  P.E = mgh  Here m = 10, g = 10m/s 2 and h = 20m.  P.E = mgh  ⇒ P.E = (10)(10)(20)  ⇒ P.E = 2000J  Now, this energy is converted completely into KE.  KE = PE  ⇒2000 =  [Tex]\frac{1}{2}mv^2[/Tex] Given m = 10Kg,  ⇒2000 =  [Tex]\frac{1}{2}10v^2[/Tex] ⇒400 = v 2 v = 20m/s 

Question 9: Suppose a rock of 100Kg was kept at 80m height. Now, this block is dropped. Find out the velocity of the block just before it hits the ground.

The block of 10Kg is kept at a height of 80m.  The potential energy of the block will be,  P.E = mgh  Here m = 100, g = 10m/s 2 and h = 80m.  P.E = mgh  ⇒ P.E = (100)(10)(80)  ⇒ P.E = 80000J  Now, this energy is converted completely into KE.  KE = PE  ⇒80000 =  [Tex]\frac{1}{2}mv^2[/Tex] Given m = 100Kg,  ⇒80000 =  [Tex]\frac{1}{2}100v^2[/Tex] ⇒1600 = v 2 v = 40m/s 

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Kinetic energy problems

When solving kinetic energy problems, you may be asked to find 3 variables. These variables are the kinetic energy, the mass, or the speed.

Problem # 1:

Suppose a car has 3000 Joules of kinetic energy. What will be its kinetic energy if the speed is doubled? What if the speed is tripled?

We already proved in kinetic energy lesson that whenever the speed is doubled, the kinetic energy is quadrupled or four times as big.

4 × 3000 = 12000

Therefore, the kinetic energy is going to be 12000 joules.

Let v be the speed of a moving object. Let speed =  3v after the speed is tripled. 

9 × 3000 = 27000

Therefore, the kinetic energy is going to be 27000 joules.

Problem # 2:

Calculate the kinetic energy of a 10 kg object moving with a speed of 5 m/s. Calculate the kinetic energy again when the speed is doubled.

Tricky kinetic energy problems

Problem # 3: 

Suppose a rat and a rhino are running with the same kinetic energy. Which one do you think is going faster?

The only tricky and hard part is to use the kinetic energy formula to solve for v.

Multiply both sides by 2

Problem # 4: 

The kinetic energy of an object is 8 times bigger than the mass. Is it possible to get speed of the object?

Think carefully and try to solve this problem yourself.

Potential energy

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Kinetic Energy: Explanation, Review, and Examples

• The Albert Team
• Last Updated On: February 16, 2023

We’ve learned a lot so far about how and why things move. What we haven’t learned is what allows one object to exert a force on another. Energy is an important concept in physics that is used to understand large and small processes. Here, we’ll begin with the energy of a moving object by exploring examples of kinetic energy and how to calculate it. We’ll also dive into the differences between kinetic and potential energy.

What We Review

What is Kinetic Energy?

“Kinetic” and “energy” are both words we should be familiar with at this point. “Kinetic” comes up when discussing the friction of a moving object. “Energy” is one you’ve probably heard a lot in your life. It probably has a meaning similar to “the ability to do something” and that’s what it means in physics, too. We refine it a bit to mean the ability to do work or exert a force on an object over a displacement. From this, we can guess that kinetic energy should have something to do with things moving and their ability to exert a force. While that isn’t a perfect definition, it is fairly close.

Kinetic energy does, indeed, have to do with an object in motion. It is the energy an object has because of its motion and we can think of it in terms of the force that can be exerted on an object or the force it would take to stop the object. The way we want to think about this force will change in different situations. In general, though, the kinetic energy of a moving object can be thought of as a measure of how far the object’s velocity is from zero. The greater the velocity, the greater the energy.

Examples of Kinetic Energy

One familiar example of kinetic energy and how it changes depending on velocity could be walking versus running. It’s much harder to stop quickly when running at full speed than it is when walking slowly. The same applies to driving a car – the faster it’s going, the sooner the brakes need to be hit. This applies to our everyday lives but is also often built into video games. Have you noticed how much harder it can be to take a sharp turn in a racing game when driving at full speed than it is when slowing down? You have kinetic energy to thank for that.

How to Calculate Kinetic Energy

So far, we’ve been talking a lot about the velocity of a moving object relative to kinetic energy. While that is an important factor, there is another factor – mass. The mass of an object is a measure of its inertia, of how hard it is to change its motion. To find the kinetic energy of an object, then, we must account for both its velocity and its mass. These two values don’t have the same impact on the kinetic energy of a moving object, though.

Kinetic Energy Formula

The formula for kinetic energy is written as:

Here, the E_{K} symbol is used to represent kinetic energy, m represents the mass, and lastly v represents the velocity. We’ll work with this equation in just a moment, but first, there are two very important conceptual points to be made here. First, is that we haven’t seen this combination of values before, which means we have a new unit to work with. Energy is measured in Joules, which is denoted J. Second, the velocity is squared while the mass is not.

How Mass and Velocity Affect Kinetic Energy

Because the velocity has a higher exponent than the mass does, the velocity will have a larger impact on the kinetic energy of a moving object. For example, if the mass were to double the kinetic energy would also double, but if the velocity doubled the kinetic energy would be quadrupled. Similarly, if the mass were halved then the kinetic energy would be halved, but if the velocity were halved the kinetic energy would be quartered.

Examples of Calculating Kinetic Energy

Now that we know the equation, we can start using it to calculate kinetic energy. We’ll follow our normal problem-solving method to first find kinetic energy and then to find mass from kinetic energy and velocity.

Example 1: Finding Kinetic Energy from Mass and Velocity

A 6\text{ kg} bowling ball is rolling toward the pins at 7\text{ m/s} . What is the kinetic energy of the bowling ball?

• m=6\text{ kg}
• v=7\text{ m/s}
• E_{K}=\text{?}
• E_{K}=\frac{1}{2}mv^{2}

Example 2: Finding Mass From Kinetic Energy and Velocity

A flamboyance of flamingo flies at an average speed of about 17\text{ m/s} . If one of the flamingos has a kinetic energy of 430\text{ J} , what is its mass?

• v=17\text{ m/s}
• E_{K}=430\text{ J}

Differences Between Kinetic and Potential Energy

Kinetic energy is not the only important energy type in mechanical physics. The other energy you’re likely to encounter is potential energy. Both kinetic energy and potential energy are important to how we view and understand the world and there is one key difference between them.

What is Potential Energy?

Potential energy is the energy stored in an object that gives it the potential to do something. This is the energy in an object that is not moving or exerting a force, but it could. We often see this as gravitational potential energy – the potential to fall – or as spring potential energy – the potential to launch something with a spring. A more in-depth explanation of potential energy can be found in a separate post , but the important thing to remember is that potential energy is the potential to complete an action.

How Kinetic and Potential Energy Relate to Each Other

Kinetic energy is energy that comes from motion. Potential energy, on the other hand, is the energy an object has the potential to use for motion. These energies can be transformed into one another. Let’s consider an example of potential energy transforming into kinetic energy. Before jumping out of a plane, a skydiver has no kinetic energy but does have a lot of potential energy. Once they jump out of the plane, their potential energy will decrease as they get closer to the ground and that energy will be shifting into kinetic energy as the skydiver’s velocity increases. The skydiver’s kinetic energy will increase at the exact same rate that its potential energy decreases. The same would be true for any object whose potential energy is transforming into kinetic energy or for one whose kinetic energy is transforming into potential energy.

Mechanical energy is constantly present in our lives and it’s an important part of how we understand the universe. Kinetic energy can be found in any moving object and there are examples of kinetic energy in many aspects of daily life. It is far from the only type of mechanical energy you’ll learn about on your physics journey, but knowing how to identify and calculate kinetic energy will help as you continue to expand your knowledge of how and why objects behave as they do.

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7.2 Kinetic Energy and the Work-Energy Theorem

Learning objectives.

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

• Explain work as a transfer of energy and net work as the work done by the net force.
• Explain and apply the work-energy theorem.

Work Transfers Energy

What happens to the work done on a system? Energy is transferred into the system, but in what form? Does it remain in the system or move on? The answers depend on the situation. For example, if the lawn mower in Figure 7.2 (a) is pushed just hard enough to keep it going at a constant speed, then energy put into the mower by the person is removed continuously by friction, and eventually leaves the system in the form of heat transfer. In contrast, work done on the briefcase by the person carrying it up stairs in Figure 7.2 (d) is stored in the briefcase-Earth system and can be recovered at any time, as shown in Figure 7.2 (e). In fact, the building of the pyramids in ancient Egypt is an example of storing energy in a system by doing work on the system. Some of the energy imparted to the stone blocks in lifting them during construction of the pyramids remains in the stone-Earth system and has the potential to do work.

In this section we begin the study of various types of work and forms of energy. We will find that some types of work leave the energy of a system constant, for example, whereas others change the system in some way, such as making it move. We will also develop definitions of important forms of energy, such as the energy of motion.

Net Work and the Work-Energy Theorem

We know from the study of Newton’s laws in Dynamics: Force and Newton's Laws of Motion that net force causes acceleration. We will see in this section that work done by the net force gives a system energy of motion, and in the process we will also find an expression for the energy of motion.

Let us start by considering the total, or net, work done on a system. Net work is defined to be the sum of work on an object. The net work can be written in terms of the net force on an object. F net F net . In equation form, this is W net = F net d cos θ W net = F net d cos θ where θ θ is the angle between the force vector and the displacement vector.

Figure 7.3 (a) shows a graph of force versus displacement for the component of the force in the direction of the displacement—that is, an F cos θ F cos θ vs. d d graph. In this case, F cos θ F cos θ is constant. You can see that the area under the graph is F d cos θ F d cos θ , or the work done. Figure 7.3 (b) shows a more general process where the force varies. The area under the curve is divided into strips, each having an average force ( F cos θ ) i ( ave ) ( F cos θ ) i ( ave ) . The work done is ( F cos θ ) i ( ave ) d i ( F cos θ ) i ( ave ) d i for each strip, and the total work done is the sum of the W i W i . Thus the total work done is the total area under the curve, a useful property to which we shall refer later.

Net work will be simpler to examine if we consider a one-dimensional situation where a force is used to accelerate an object in a direction parallel to its initial velocity. Such a situation occurs for the package on the roller belt conveyor system shown in Figure 7.4 .

The force of gravity and the normal force acting on the package are perpendicular to the displacement and do no work. Moreover, they are also equal in magnitude and opposite in direction so they cancel in calculating the net force. The net force arises solely from the horizontal applied force F app F app and the horizontal friction force f f . Thus, as expected, the net force is parallel to the displacement, so that θ = 0º θ = 0º and cos θ = 1 cos θ = 1 , and the net work is given by

The effect of the net force F net F net is to accelerate the package from v 0 v 0 to v v . The kinetic energy of the package increases, indicating that the net work done on the system is positive. (See Example 7.2 .) By using Newton’s second law, and doing some algebra, we can reach an interesting conclusion. Substituting F net = ma F net = ma from Newton’s second law gives

To get a relationship between net work and the speed given to a system by the net force acting on it, we take d = x − x 0 d = x − x 0 and use the equation studied in Motion Equations for Constant Acceleration in One Dimension for the change in speed over a distance d d if the acceleration has the constant value a a ; namely, v 2 = v 0 2 + 2 ad v 2 = v 0 2 + 2 ad (note that a a appears in the expression for the net work). Solving for acceleration gives a = v 2 − v 0 2 2 d a = v 2 − v 0 2 2 d . When a a is substituted into the preceding expression for W net W net , we obtain

The d d cancels, and we rearrange this to obtain

This expression is called the work-energy theorem , and it actually applies in general (even for forces that vary in direction and magnitude), although we have derived it for the special case of a constant force parallel to the displacement. The theorem implies that the net work on a system equals the change in the quantity 1 2 mv 2 1 2 mv 2 . This quantity is our first example of a form of energy.

The Work-Energy Theorem

The net work on a system equals the change in the quantity 1 2 mv 2 1 2 mv 2 .

The quantity 1 2 mv 2 1 2 mv 2 in the work-energy theorem is defined to be the translational kinetic energy (KE) of a mass m m moving at a speed v v . ( Translational kinetic energy is distinct from rotational kinetic energy, which is considered later.) In equation form, the translational kinetic energy,

is the energy associated with translational motion. Kinetic energy is a form of energy associated with the motion of a particle, single body, or system of objects moving together.

We are aware that it takes energy to get an object, like a car or the package in Figure 7.4 , up to speed, but it may be a bit surprising that kinetic energy is proportional to speed squared. This proportionality means, for example, that a car traveling at 100 km/h has four times the kinetic energy it has at 50 km/h, helping to explain why high-speed collisions are so devastating. We will now consider a series of examples to illustrate various aspects of work and energy.

Example 7.2

Calculating the kinetic energy of a package.

Suppose a 30.0-kg package on the roller belt conveyor system in Figure 7.4 is moving at 0.500 m/s. What is its kinetic energy?

Because the mass m m and speed v v are given, the kinetic energy can be calculated from its definition as given in the equation KE = 1 2 mv 2 KE = 1 2 mv 2 .

The kinetic energy is given by

Entering known values gives

which yields

Note that the unit of kinetic energy is the joule, the same as the unit of work, as mentioned when work was first defined. It is also interesting that, although this is a fairly massive package, its kinetic energy is not large at this relatively low speed. This fact is consistent with the observation that people can move packages like this without exhausting themselves.

Example 7.3

Determining the work to accelerate a package.

Suppose that you push on the 30.0-kg package in Figure 7.4 with a constant force of 120 N through a distance of 0.800 m, and that the opposing friction force averages 5.00 N.

(a) Calculate the net work done on the package. (b) Solve the same problem as in part (a), this time by finding the work done by each force that contributes to the net force.

Strategy and Concept for (a)

This is a motion in one dimension problem, because the downward force (from the weight of the package) and the normal force have equal magnitude and opposite direction, so that they cancel in calculating the net force, while the applied force, friction, and the displacement are all horizontal. (See Figure 7.4 .) As expected, the net work is the net force times distance.

Solution for (a)

The net force is the push force minus friction, or F net = 120 N – 5 . 00 N = 115 N F net = 120 N – 5 . 00 N = 115 N . Thus the net work is

Discussion for (a)

This value is the net work done on the package. The person actually does more work than this, because friction opposes the motion. Friction does negative work and removes some of the energy the person expends and converts it to thermal energy. The net work equals the sum of the work done by each individual force.

Strategy and Concept for (b)

The forces acting on the package are gravity, the normal force, the force of friction, and the applied force. The normal force and force of gravity are each perpendicular to the displacement, and therefore do no work.

Solution for (b)

The applied force does work.

The friction force and displacement are in opposite directions, so that θ = 180º θ = 180º , and the work done by friction is

So the amounts of work done by gravity, by the normal force, by the applied force, and by friction are, respectively,

The total work done as the sum of the work done by each force is then seen to be

Discussion for (b)

The calculated total work W total W total as the sum of the work by each force agrees, as expected, with the work W net W net done by the net force. The work done by a collection of forces acting on an object can be calculated by either approach.

Example 7.4

Determining speed from work and energy.

Find the speed of the package in Figure 7.4 at the end of the push, using work and energy concepts.

Here the work-energy theorem can be used, because we have just calculated the net work, W net W net , and the initial kinetic energy, 1 2 m v 0 2 1 2 m v 0 2 . These calculations allow us to find the final kinetic energy, 1 2 mv 2 1 2 mv 2 , and thus the final speed v v .

The work-energy theorem in equation form is

Solving for 1 2 mv 2 1 2 mv 2 gives

Solving for the final speed as requested and entering known values gives

Using work and energy, we not only arrive at an answer, we see that the final kinetic energy is the sum of the initial kinetic energy and the net work done on the package. This means that the work indeed adds to the energy of the package.

Example 7.5

Work and energy can reveal distance, too.

How far does the package in Figure 7.4 coast after the push, assuming friction remains constant? Use work and energy considerations.

We know that once the person stops pushing, friction will bring the package to rest. In terms of energy, friction does negative work until it has removed all of the package’s kinetic energy. The work done by friction is the force of friction times the distance traveled times the cosine of the angle between the friction force and displacement; hence, this gives us a way of finding the distance traveled after the person stops pushing.

The normal force and force of gravity cancel in calculating the net force. The horizontal friction force is then the net force, and it acts opposite to the displacement, so θ = 180º θ = 180º . To reduce the kinetic energy of the package to zero, the work W fr W fr by friction must be minus the kinetic energy that the package started with plus what the package accumulated due to the pushing. Thus W fr = − 95 . 75 J W fr = − 95 . 75 J . Furthermore, W fr = f d ′ cos θ = – f d ′ W fr = f d ′ cos θ = – f d ′ , where d ′ d ′ is the distance it takes to stop. Thus,

This is a reasonable distance for a package to coast on a relatively friction-free conveyor system. Note that the work done by friction is negative (the force is in the opposite direction of motion), so it removes the kinetic energy.

Some of the examples in this section can be solved without considering energy, but at the expense of missing out on gaining insights about what work and energy are doing in this situation. On the whole, solutions involving energy are generally shorter and easier than those using kinematics and dynamics alone.

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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/college-physics-2e/pages/1-introduction-to-science-and-the-realm-of-physics-physical-quantities-and-units

• Authors: Paul Peter Urone, Roger Hinrichs
• Publisher/website: OpenStax
• Book title: College Physics 2e
• Publication date: Jul 13, 2022
• Location: Houston, Texas
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• Section URL: https://openstax.org/books/college-physics-2e/pages/7-2-kinetic-energy-and-the-work-energy-theorem

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• Physics Formulas

Kinetic Energy Formula

Kinetic energy is the energy possessed by a body due to its motion. Kinetic Energy Formula is articulated as

mass of the body = m ,

the velocity with which the body is travelling is v .

The Kinetic energy is articulated in Kgm 2 /s 2

Kinetic energy formula is used to compute the mass, velocity or kinetic energy of the body if any of the two numerics are given.

Kinetic Energy Solved Examples

Underneath are questions on Kinetic energy which aids one to understand where they can use these questions.

Problem 1:  A car is travelling at a velocity of 10 m/s and it has a mass of 250 Kg. Compute its Kinetic energy? Answer:

Given: Mass of the body m = 250 Kg, Velocity v = 10 m/s,

Kinetic energy is given by

=12500 kgm 2 s 2

Problem  2:  A man is transporting a trolley of mass 6 Kg and having Kinetic energy of 40 J. Compute its Velocity with which he is running? Answer:

Given: Mass, m = 6 Kg Kinetic energy K.E = 60 J

The man is running with the velocity of 3.65m/s

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the maximum speed of “tsunami” waves is 100m/s .calculate the maximum kinetic energy of every kilogram of water in these “tsunami “.

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SOLVING PROBLEMS ON KINETIC ENERGY

Formula to find kinetic energy of an object :

K.E = (1/2)mv 2

where 'm' is the mass of the object and 'v' is its velocity.

Kinetic energy of an object is measured in joules.

Example 1 :

What is the Kinetic Energy of a 150 kg object that is moving with a speed of 15 m/s?

Substitute m = 150 and v = 15.

= (1/2)(150)(15) 2

= (1/2)(150)(225)

The Kinetic Energy of the object is 16875 joules.

Example 2 :

An object has a kinetic energy of 425 joules and a mass of 34 kg. How fast is the object moving?

K.E = 425 joules

(1/2)mv 2 = 425

Substitute K. E = 25 and m = 34.

(1/2)(34)v 2  = 425

17v 2  = 425

Divide both sides by 17.

v 2  = 25

Take square root on both sides.

v   = 5

The object is moving at a rate of 5 m/s.

Example 3 :

An object moving with a speed of 25 m/s and has a kinetic energy of 1875 joules. What is the mass of the object?

K.E = 1875 joules

(1/2)mv 2  = 1875

Substitute v = 25.

(1/2)m(25) 2  = 1875

625m/2   = 1875

Multiply both sides by 2.

625m = 3750

Divide both sides by 625.

The mass of the object is 6 kg.

Example 4 :

If the mass of an object is halved and its speed is doubled, how does the kinetic energy change?

Let m be the mass and v be the speed of the object.

Then, the kinetic energy is

Since the mass of is halved and its speed is doubled, replace m by (1/2)m and v by 2v.

New K.E = (1/2)(1/2)m(2v) 2

= (1/2)(1/2)m(4v 2 )

= (1/2)m(2v 2 )

= 2 x (1/2)mv 2

If the mass of an object is halved and its speed is doubled, the kinetic energy is doubled.

Example 5 :

If the kinetic energy of a moving tennis ball is doubled, its velocity must have increased by what factor?

Formula for kinetic energy.

When the kinetic energy is doubled, let us assume that the velocity has to be increased by the factor 'x'.

2K.E = (1/2)m(xv) 2

2K.E = (1/2)mx 2 v 2

2K.E = x 2 (1/2)mv 2

2K.E = x 2 K.E

Divide both sides by K.E.

If the kinetic energy of a moving tennis ball is doubled, its velocity must have increased by the factor  √2.

Example 6 :

A radioactive element loses 15 percent of its mass and 20 percent of its velocity. By what percent has its kinetic energy decreased?

Since radioactive element loses 15 percent of its mass and 20 percent of its velocity, replace 'm' by '0.85m' and 'v' by '0.8v'.

New K.E = (1/2)(0.85m)(0.8v) 2

= (1/2)(0.85m)(0.64v 2 )

= 0.544 x (1/2)mv 2

= 0.544 x K.E

= 54.4 % of K.E

After  15 percent loss in mass and 20 percent loss velocity, the kinetic energy becomes 54.4% of its original level.

100% - 54.4% = 45.6%

After the radioactive element loses 15 percent of its mass and 20 percent of its velocity, the kinetic energy has decreased by 45.6%

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Kinetic Energy

Now, the bears I live with average, the males, eight to twelve hundred pounds [360 to 540 kg]. They're the largest bears in the world…. They've been clocked at 41 [mph] and they've run a hundred meter dash in 5.85 seconds, which a human on steroids doesn't even approach. Timothy Treadwell, 2001

• Compute the speed of a grizzly bear using Mr. Treadwell's hundred meter statement.
• Compute the kinetic energy of a grizzly bear using the speed you calculated in part a. and the average mass stated by Mr. Treadwell.
• How fast would a 250 lb man have to run to have the same kinetic energy you calculated in part b? (Do not use a calculator to compute your answer.)
• How fast would a 4000 lb car have to drive to have the same kinetic energy you calculated in part b? (Do not use a calculator to compute your answer.)

The Space Shuttle Columbia disintegrated during reentry on the morning of 1 February 2003. The cause of the accident was determined months later. A review of video footage taken during the launch 16 days earlier showed a large piece of foam insulation falling off the external fuel tank shortly after liftoff then striking the leading edge of the orbiter's left wing. This compromised the thermal protection system at the point of impact and allowed the superheated gases generated on reentry to melt the aluminum frame there. The left wing snapped off first, the orbiter tumbled and broke apart, scattering pieces across eastern Texas. All seven crew onboard were killed

Eighty-two seconds into STS 107 [the mission number], a sizeable piece of debris struck the left wing of the Columbia. Visual evidence and other sensor data established that the debris came from the bipod ramp area and impacted the wing on the wing leading edge. At this time Columbia was traveling at a speed of about 2,300 feet/second (fps) through an altitude of about 65,900 feet. Based on a combination of image analysis and advanced computational methods, the Board determined that a foam projectile with a total weight of 1.67 lb and impact velocity of 775 fps would best represent the debris strike…. Just prior to separating from the External Tank (ET), the foam was traveling with the orbiter at about 2,300 fps. The visual evidence shows that the debris impacted the wing approximately 0.161 seconds after separating from the ET. In that time, the debris slowed down from 2,300 fps to about 1,500 fps, so it hit the orbiter with a relative velocity of about 800 fps. In essence, the debris slowed down and the Orbiter did not, so that the Orbiter ran into the debris. Columbia Accident Investigation Board, 2003

Show that a piece of rigid foam insulation like the one that struck the Space Shuttle Columbia possesses a considerable amount of kinetic energy despite being "just a piece of foam".

• Determine the kinetic energy of the foam debris that struck Columbia in 2003.
• How fast would a 10 lb sledge hammer have to travel in order to have the same kinetic energy as the foam? State your answer in miles per hour or kilometers per hour as you prefer.
• How massive would a defensive tackle of American or Canadian football have to be if he ran as fast as a world class sprinter and had the same kinetic energy as the foam debris? State your answer in pounds or kilograms as you prefer.
• Write something different.
• Write something completely different.
• Is it possible for a motorcycle to have more kinetic energy than a truck?

Verify Robinson's first law of space combat (originally known as Robinson's first law of science fiction).

An object impacting at 3 km/s delivers kinetic energy equal to its mass in TNT. Ken Burnside, 2003

The term energy may be applied, with great propriety, to the product of the mass or weight of a body, into the square of the number expressing its velocity. Thus, if a weight of one ounce moves with the velocity of a foot in a second, we may call its energy 1; if a second body of two ounces have a velocity of three feet in a second, its energy will be twice the square of three, or 18. Thomas Young, 1807

Young would not receive full credit on an exam were he a student today. He provided units for the quantitites used in his calculations, but he neglected to include them in his solutions. Let young be the name of the unit that is missing from the passage above.

• How many youngs are in a joule (the unit of energy in the International System )?
• How many youngs are in a foot-pound (the unit of energy in the British Engineering System )?
• How many ergs (the unit of energy in the Gaussian System ) are in a young?

• tons of TNT (For comparison, the largest nuclear weapon ever tested had a yield of 50 million tons of TNT.)

Assuming that intensity is based on the kinetic energy of a "piece" of moving air, how many times more intense is…

• an EF2 than an EF1 tornado,
• an EF5 than an EF4 tornado,
• an EF5 than an EF1 tornado?

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What Is Kinetic Energy? Kinetic Energy Examples

Potential and kinetic energy are the two major types of energy . Here is a look at kinetic energy, including its definition, examples, units, formula, and how to calculate it.

Kinetic Energy Definition

In physics, kinetic energy is the energy an object has due to its motion. It is defined as the work required to accelerate a body of a given mass from rest to a certain velocity. Once the mass reaches the velocity, its kinetic energy remains unchanged unless its speed changes. However, velocity and thus kinetic energy depend on the frame of reference. In other words, an object’s kinetic energy is not invariant.

Kinetic Energy Units

The SI unit of kinetic energy is the joule (J), which is a kg⋅m 2 ⋅s −2 . The English unit of kinetic energy is the foot-pound (ft⋅lb). Kinetic energy is a scalar quantity . It has magnitude, but no direction.

Kinetic Energy Examples

Anything you can think of that has mass (or apparent mass) and motion is an example of kinetic energy. Kinetic energy examples include:

• A flying aircraft, bird, or superhero
• Walking, jogging, bicycling, swimming, dancing, or running
• Falling down or dropping an object
• Throwing a ball
• Driving a car
• Playing with a yo-yo
• Launching a rocket
• A windmill spinning
• Clouds moving across the sky
• An avalanche
• A waterfall or flowing stream
• Electricity flowing through a wire
• Orbiting satellites
• A meteor falling to Earth
• Sound moving from a speaker to your ears
• Electrons orbiting the atomic nucleus
• Light traveling from the Sun to the Earth (photons have momentum, so they have apparent mass)

Kinetic Energy Formula

The formula for kinetic energy (KE) relates energy to mass (m) and velocity (v).

KE = 1/2 mv 2

Because mass is always a positive value and the square of any value is a positive number, kinetic energy is always positive. Also, this means the maximum kinetic energy occurs when velocity is greatest, regardless of the direction of motion.

From the kinetic energy equation, you can see an object’s velocity matters more than its mass. So, even a small object has a lot of kinetic energy if it’s moving quickly.

The kinetic energy formula works in classical physics, but it starts to deviate from true energy when the velocity approaches the speed of light ( c ).

How to Calculate Kinetic Energy

The key to solving kinetic energy problems is to remember that 1 joule equals 1 kg⋅m 2 ⋅s −2 . Speed is the magnitude of velocity, so you can use it in the kinetic energy equation. Otherwise, watch your units in fractions. For example, (1)/(400 m 2 /s 2 ) is the same as (1/400) s 2 /m 2 .

Calculate the kinetic energy of a 68 kg person moving with a speed of 1.4 m/s (in other words, the kinetic energy of a typical person walking).

Plugging in the numbers:

KE = 1/2(68 kg)(1.4 m/s) 2 KE = 66.64 kg⋅m 2 ⋅s −2 KE = 66.64 J

Calculate the mass of an object moving at 20 m/s with a kinetic energy of 1000 J.

Rearrange the kinetic energy equation to solve for mass:

m = 2KE/v 2 m = (2)(1000 kg⋅m 2 ⋅s −2 )/(20 m/s) 2 m= (2000 kg⋅m 2 ⋅s −2 )/(400 m 2 /s 2 ) m = 5 kg

Difference Between Kinetic and Potential Energy

Kinetic energy can transform into potential energy , and vice versa. Kinetic energy is the energy associated with a body’s motion, while potential energy is the energy due to an object’s position. All the other types of energy (e.g., electrical energy , chemical energy , thermal energy, nuclear energy) have kinetic energy, potential energy, or a combination of the two. The sum of the kinetic and potential energy of a system (its total energy) is a constant because of Conservation of Energy . In quantum mechanics, the sum of kinetic and potential energy is called the Hamiltonian.

A frictionless roller coaster is a good example of the interplay between kinetic and potential energy . At the top of the track, the roller coaster has maximum potential energy, but minimum kinetic energy (zero). As the cart goes down the track, its velocity increases. At the bottom of the track, the potential energy is at its minimum (zero), while the kinetic energy is at its maximum.

• Goel, V. K. (2007). Fundamentals Of Physics . Tata McGraw-Hill Education. ISBN 978-0-07-062060-5.
• Serway, Raymond A.; Jewett, John W. (2004). Physics for Scientists and Engineers (6th ed.). Brooks/Cole. ISBN 0-534-40842-7.
• Tipler, Paul; Llewellyn, Ralph (2002). Modern Physics (4th ed.). W. H. Freeman. ISBN 0-7167-4345-0.

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Mechanical Energy Problem Solutions

Mechanical energy problems and solutions.

See examples of mechanical energy problems involving kinetic energy, potential energy, and the conservation of energy. Check your work with ours.

1. How much gravitational potential energy do you have when you lift a 15 N object 10 meters off the ground?

2. How much gravitational potential energy is in a 20 kg mass when 0.6 meters above the ground?

3. How much gravitational potential energy does a 35 kg boulder have when 30 meters off the ground?

4. How many times greater is an objects potential energy when three times higher?

If you need help on ratio problems click the link below:

Rule of Ones: analyzing equations to determine how other variables change

5.  How much kinetic energy does a 0.15 kg ball thrown at 24 m/s have?

6.  How many times greater is the kinetic energy of a ball that is going five times faster?

7.  How much kinetic energy does a 1.2 kg ball have the moment it hits the ground 3.5 meters below when it starts from rest?

I cancelled out the initial kinetic energy because:

• KE i = ½ mv f 2
• KE i = (½)(3.5)(0 2 ) = 0 J

I cancelled out the final potential energy because:

• PE f = mgh f
• PE f = (3.5)(9.8)(0) = 0 J

8.  How fast is a 1.2 kg ball traveling the moment it hits the ground 3.5 meters below when it starts from rest?

(Note: In many of these problems I could cancel out mass but did not since it was provided)

Since I did not cancel out mass I could answer the following questions if asked:

• How much mechanical energy did you have at the beginning? (41.6 J)
• How much kinetic energy did you have at the beginning? (0 J)
• How much potential energy did you have at the beginning? (41.6 J)
• How much potential energy do you have at the end? (0 J)

If I cancelled out mass in my work it would not show the actual initial potential energy since PE i = mgh and not just gh.

9.  A 3.5 kg ball fell from a height of 12 meters.  How fast is it traveling when its still 5 meters off the ground?

10. An 85kg roller coaster cart is traveling 4 m/s at the top of a hill 50 meters off the ground.  How fast is it traveling at top of a second hill 20 meters off the ground?

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Potential and Kinetic Energy

Energy is the capacity to do work .

The unit of energy is J (Joule) which is also kg m 2 /s 2 (kilogram meter squared per second squared)

Energy can be in many forms! Here we look at Potential Energy (PE) and Kinetic Energy (KE).

Potential Energy and Kinetic Energy

• when raised up has potential energy (the energy of position or state)
• when falling down has kinetic energy (the energy of motion)

Potential energy (PE) is stored energy due to position or state

• a raised hammer has PE due to gravity.
• fuel and explosives have Chemical PE
• a coiled spring or a drawn bow also have PE due to their state

Kinetic energy (KE) is energy of motion

From PE to KE

For a good example of PE and KE have a play with a pendulum .

Gravitational Potential Energy

When the PE is due to an objects height then:

PE due to gravity = m g h

• m is the objects mass (kg)
• g is the "gravitational field strength" of 9.8 m/s 2 near the Earth's surface
• h is height (m)

Example: This 2 kg hammer is 0.4 m up. What is it's PE?

Kinetic energy.

The formula is:

KE = ½ m v 2

• m is the object's mass (kg)
• v is the object's speed (m/s)

Example: What is the KE of a 1500 kg car going at suburban speed of 14 m/s (about 50 km/h or 30 mph)?

KE = ½ × 1500 kg × (14 m/s) 2

KE = 147,000 kg m 2 /s 2

KE = 147 kJ

Let's double the speed!

Example: The same car is now going at highway speed of 28 m/s (about 100 km/h or 60 mph)?

KE = ½ × 1500 kg × (28 m/s) 2

KE = 588,000 kg m 2 /s 2

KE = 588 kJ

Wow! that is a big increase in energy! Highway speed is way more dangerous.

Double the speed and the KE increases by four times. Very important to know

A 1 kg meteorite strikes the Moon at 11 km/s. How much KE is that?

KE = ½ × 1 kg × (11,000 m/s) 2

KE = 60,500,000 J

KE = 60.5 MJ

That is 100 times the energy of a car going at highway speed.

When falling, an object's PE due to gravity converts into KE and also heat due to air resistance.

Let's drop something!

Example: We drop this 0.1 kg apple 1 m. What speed does it hit the ground with?

At 1 m above the ground it's Potential Energy is

PE = 0.1 kg × 9.8 m/s 2 × 1 m

PE = 0.98 kg m 2 /s 2

Ignoring air resistance (which is small for this little drop anyway) that PE gets converted into KE:

Swap sides and rearrange:

½ m v 2 = KE

v 2 = 2 × KE / m

v = √( 2 × KE / m )

Now put PE into KE and we get:

v = √( 2 × 0.98 kg m 2 /s 2 / 0.1 kg )

v = √( 19.6 m 2 /s 2 )

v = 4.427... m/s

Note: for velocity we can combine the formulas like this:

The mass does not matter! It is all about height and gravity. For our earlier example:

v = √( 2gh )

v = √( 2 × 9.8 m/s 2 × 1 m )

• Energy is the ability to do work

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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.

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.

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View Audio Guided Solution for Problem:

Kinetic Energy Examples

A series of free GCSE/IGCSE Physics Notes and Lessons .

In these lessons, we will

• Describe what is meant by kinetic energy.
• Calculate kinetic energy for a moving object.

Related Pages Energy Transfers Mechanical, Potential and Kinetic Energy Elastic Potential Energy Lessons for IGCSE Physics

Kinetic Energy

The following diagram shows the formula for kinetic energy. Scroll down the page for more examples and solutions on how to use the formula.

Kinetic energy is the energy stored in moving objects. Stationary objects have no kinetic energy.

E k = 0.5 × m × v 2

• A car with a mass of 700 kg is moving with a speed of 20m/s. Calculate the kinetic energy of the car.
• A cyclist and bike have a total mass of 100 kg and a speed of 15 m/s. Calculate the kinetic energy.
• A tennis ball is traveling at 50 m/s and has a kinetic energy of 75 J. Calculate the mass of the tennis ball.

Kinetic Energy - IGCSE Physics

Calculations using the kinetic energy formula.

• A 30 gram bullet travels at 300 m/s. How much kinetic energy does it have?
• A 70kg man runs at a pace of 4 m/s and a 50g meteor travels at 2 km/s. Which has the most kinetic energy?.
• A 5000kg truck has 400000J of kinetic energy. How fast is it moving?
• A car traveling at 10 m/s has 200000J of kinetic energy. What is the mass of the car?

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Work and kinetic energy – problems and solutions

Work-Kinetic energy :

Final speed (v t ) = 20 m/s

W net = net work

W net = ½ m (v t 2 – v o 2 )

W net = ½ (5000)(20 2 – 0 2 )

Initial speed (v o ) = 5 m/s

W net = (5)(75)

Final speed (v t ) = 5 m/s

W net = ½ (2000)(5 2 – 10 2 )

(1) Work done on the object is 30,240 Joule

Force (F) = 60 N

Acceleration of object :

v t = 6 + (6)(12)

s = 72 + (3)(144)

KE = 1/2 m v t 2 = 1/2 (10)(78) 2 = (5)(6084) = 30,420 Joule

The larger work :

6. A 4000-kg car travels along straight line at 25 m/s. The car is decelerated so that the car’s final velocity is 15 m/s. What is the work done on the car.

The change in height (h) = 5 m – 2 m = 3 meters

PE = m g h = (0.1)(10)(3) = 3 Joule.

Initial velocity (v o ) = 0

W net = ½ (1000)(5 2 – 0 2 ) = (500)(25 – 0) = (500)(25) = 12,500 Joule

Final velocity (v t ) = 0 (velocity at the highest point)

Wnet = ½ m v t 2 – ½ m v o 2 = ½ m (v t 2 – v o 2 )

10. A 1-kg object free fall with the height difference = 2.5 meters. Acceleration due to gravity is 10 m.s -2 . What is the work done on the object?

Wanted : Net work during displacement

Initial speed of car (v o ) = 36 km/hour = 36,000 meters / 3600 second = 10 meters/second

W = (1500)(150)

Initial speed (v o ) = 72 km/hour = 72,000 meters / 3600 second = 20 m/s

Theorem of work-kinetic energy states that the net work acts on an object same as the change of the kinetic energy of the object.

W net = ½ m v t 2 – ½ m v o 2

W net = ½ (2)(40 2 – 20 2 )

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