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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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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

Substituting the value of “d” in the equation, 

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

In case, initial velocity u = 0, 

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 =  K.E =  ⇒ K.E =  ⇒ 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 =  K.E =  ⇒ K.E =  ⇒ 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 =  K.E =  ⇒ K.E =  ⇒ 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.

Given: W = 100J Work done by the force is equal to the change in kinetic energy.  W =  Given, u = 2 m/s and v = ?, m = 2kg.  Plugging the values in the given equation,  W =  ⇒  ⇒

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 =  Given, u = 10m/s and v = ? . m = 2kg.  Plugging the values in the given equation,  W =  ⇒  ⇒  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 =    KE of the heavier object  M =1000Kg and v = 10m/s  K.E =    ⇒ K.E =  ⇒K.E = 50,000J  Now this energy is transferred to another ball.  m = 10Kg and v = ? 50,000 =  ⇒ 10,000 = v 2 ⇒ v = 100 m/s

Question 7: Suppose a 10Kg 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? 

KE is given by the formula, K.E =    KE of the heavier object  M =10Kg and v = 100m/s  K.E =    ⇒ K.E =  ⇒K.E = 50,000J  Now this energy is transferred to another ball.  m = 20Kg and v = ? 50,000 =  ⇒ 5000 = v 2 ⇒ v = 50√2 m/s

Question 8: Suppose a 10Kg 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? 

Question 9: Suppose a 10Kg 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 =  Given m = 10Kg,  ⇒2000 =  ⇒400 = v 2 v = 20m/s 

Question 10: 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 20m.  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 =  Given m = 100Kg,  ⇒80000 =  ⇒1600 = v 2 v = 40m/s 

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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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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?

investigative

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Middle school physics - NGSS

Course: middle school physics - ngss   >   unit 3.

• Kinetic energy

Understand: kinetic energy

• (Choice A)   Louie has a quarter the kinetic energy of Alex. A Louie has a quarter the kinetic energy of Alex.
• (Choice B)   Louie has half the kinetic energy of Alex. B Louie has half the kinetic energy of Alex.
• (Choice C)   Louie has twice the kinetic energy of Alex. C Louie has twice the kinetic energy of Alex.
• (Choice D)   Louie has four times the kinetic energy of Alex. D Louie has four times the kinetic energy of Alex.

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Potential And Kinetic Energy Example Problem – Work and Energy Examples

Potential energy is energy attributed to an object by virtue of its position. When the position is changed, the total energy remains unchanged but some potential energy gets converted into kinetic energy . The frictionless roller coaster is a classic potential and kinetic energy example problem.

The roller coaster problem shows how to use the conservation of energy to find the velocity or position or a cart on a frictionless track with different heights. The total energy of the cart is expressed as a sum of its gravitational potential energy and kinetic energy. This total energy remains constant across the length of the track.

Potential And Kinetic Energy Example Problem

A cart travels along a frictionless roller coaster track. At point A, the cart is 10 m above the ground and traveling at 2 m/s. A) What is the velocity at point B when the cart reaches the ground? B) What is the velocity of the cart at point C when the cart reaches a height of 3 m? C) What is the maximum height the cart can reach before the cart stops?

The total energy of the cart is expressed by the sum of its potential energy and its kinetic energy.

Potential energy of an object in a gravitational field is expressed by the formula

where PE is the potential energy m is the mass of the object g is the acceleration due to gravity = 9.8 m/s 2 h is the height above the measured surface.

Kinetic energy is the energy of the object in motion. It is expressed by the formula

KE = ½mv 2

where KE is the kinetic energy m is the mass of the object v is the velocity of the object.

The total energy of the system is conserved at any point of the system. The total energy is the sum of the potential energy and the kinetic energy.

Total E = KE + PE

To find the velocity or position, we need to find this total energy. At point A, we know both the velocity and the position of the cart.

Total E = KE + PE Total E = ½mv 2  + mgh Total E = ½m(2 m/s) 2  + m(9.8 m/s 2 )(10 m) Total E = ½m(4 m 2 /s 2 ) + m(98 m 2 /s 2 ) Total E = m(2 m 2 /s 2 ) + m(98 m 2 /s 2 ) Total E = m(100 m 2 /s 2 )

We can leave the mass value as it appears for now. As we complete each part, you will see what happens to this variable.

The cart is at ground level at point B, so h = 0 m.

Total E = ½mv 2  + mgh Total E = ½mv 2  + mg(0 m) Total E = ½mv 2

All of the energy at this point is kinetic energy. Since total energy is conserved, the total energy at point B is the same as the total energy at point A.

Total E at A = Total Energy at B m(100 m 2 /s 2 ) = ½mv 2

Divide both sides by m 100 m 2 /s 2 = ½v 2

Multiply both sides by 2 200 m 2 /s 2 = v 2

v = 14.1 m/s

The velocity at point B is 14.1 m/s.

At point C, we know only a value for h (h = 3 m).

Total E = ½mv 2 + mgh Total E = ½mv 2 + mg(3 m)

As before, the total energy is conserved. Total energy at A = total energy at C.

m(100 m 2 /s 2 ) = ½mv 2 + m(9.8 m/s 2 )(3 m) m(100 m 2 /s 2 ) = ½mv 2 + m(29.4 m 2 /s 2 )

Divide both sides by m

100 m 2 /s 2 = ½v 2 + 29.4 m 2 /s 2 ½v 2 = (100 – 29.4) m 2 /s 2 ½v 2 = 70.6 m 2 /s 2 v 2 = 141.2 m 2 /s 2 v = 11.9 m/s

The velocity at point C is 11.9 m/s.

The cart will reach its maximum height when the cart stops or v = 0 m/s.

Total E = ½mv 2 + mgh Total E = ½m(0 m/s) 2 + mgh Total E = mgh

Since total energy is conserved, the total energy at point A is the same as the total energy at point D.

m(100 m 2 /s 2 ) = mgh

100 m 2 /s 2  = gh

100 m 2 /s 2  = (9.8 m/s 2 ) h

The maximum height of the cart is 10.2 m.

A) The velocity of the cart at ground level is 14.1 m/s. B) The velocity of the cart at a height of 3 m is 11.9 m/s. C) The maximum height of the cart is 10.2 m.

This type of problem has one main key point: total energy is conserved at all points of the system. If you know the total energy at one point, you know the total energy at all points.

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• Kinetic and Potential Energy of Motion

Lesson Kinetic and Potential Energy of Motion

Grade Level: 8 (7-9)

Time Required: 45 minutes

Lesson Dependency: None

Subject Areas: Physical Science, Physics

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Mechanical engineers are concerned about the mechanics of energy — how it is generated, stored and moved. Product design engineers apply the principles of potential and kinetic energy when they design consumer products. For example, a pencil sharpener employs mechanical energy and electrical energy. When designing a roller coaster, mechanical and civil engineers ensure that there is sufficient potential energy (which is converted to kinetic energy) to move the cars through the entire roller coaster ride.

After this lesson, students should be able to:

• Recognize that engineers need to understand the many different forms of energy in order to design useful products.
• Explain the concepts of kinetic and potential energy.
• Understand that energy can change from one form into another.
• Understand that energy can be described by equations.

Educational Standards Each TeachEngineering lesson or activity is correlated to one or more K-12 science, technology, engineering or math (STEM) educational standards. All 100,000+ K-12 STEM standards covered in TeachEngineering are collected, maintained and packaged by the Achievement Standards Network (ASN) , a project of D2L (www.achievementstandards.org). In the ASN, standards are hierarchically structured: first by source; e.g. , by state; within source by type; e.g. , science or mathematics; within type by subtype, then by grade, etc .

Ngss: next generation science standards - science, common core state standards - math.

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State standards, colorado - math, colorado - science.

Begin by showing the class three items: 1) an item of food (such as a bagel, banana or can of soda water), 2) a battery, and 3) you, standing on a stool or chair. Ask the class what these three things have in common. The answer is energy. The food contains chemical energy that is used by the body as fuel. The battery contains electrical energy (in the form of electrical, potential or stored energy), which can be used by a flashlight or a portable CD player. A person standing on a stool has potential energy (sometimes called gravitational potential energy) that could be used to crush a can, smash the banana, or really hurt the foot of someone standing under you. Do a dramatic demonstration of jumping down on the banana or an empty soda can. (Be careful! Banana peels are slippery!) Explain the ideas of potential energy and kinetic energy as two different kinds of mechanical energy . Give definitions of each and present the equations, carefully explaining each variable, as discussed in the next section,

PE = mass x g x height

KE = 1/2 m x v 2

Lesson Background and Concepts for Teachers

Whenever something moves, you can see the change in energy of that system. Energy can make things move or cause a change in the position or state of an object. Energy can be defined as the capacity for doing work. Work is done when a force moves an object over a given distance. The capacity for work, or energy, can come in many different forms. Examples of such forms are mechanical, electrical, chemical or nuclear energy.

This lesson introduces mechanical energy , the form of energy that is easiest to observe on a daily basis. All moving objects have mechanical energy. There are two types of mechanical energy: potential energy and kinetic energy. Potential energy is the energy that an object has because of its position and is measured in Joules (J). Potential energy can also be thought of as stored energy. Kinetic energy is the energy an object has because of its motion and is also measured in Joules (J). Due to the principle of conservation of energy, energy can change its form (potential, kinetic, heat/thermal, electrical, light, sound, etc.) but it is never created or destroyed.

Within the context of mechanical energy, potential energy is a result of an object's position, mass and the acceleration of gravity. A book resting on the edge of a table has potential energy; if you were to nudge it off the edge, the book would fall. It is sometimes called gravitational potential energy ( PE ). It can be expressed mathematically as follows:

PE = mass x g x height or PE = weight x height

where PE is the potential energy, and g is the acceleration due to gravity. At sea level, g = 9.81 meters/sec 2 or 32.2 feet/sec 2 . In the metric system, we would commonly use mass in kilograms or grams with the first equation. With English units it is common to use weight in pounds with the second equation.

Kinetic energy ( KE ) is energy of motion. Any object that is moving has kinetic energy. An example is a baseball that has been thrown. The kinetic energy depends on both mass and velocity and can be expressed mathematically as follows:

Here KE stands for kinetic energy. Note that a change in the velocity will have a much greater effect on the amount of kinetic energy because that term is squared. The total amount of mechanical energy in a system is the sum of both potential and kinetic energy, also measured in Joules (J).

Total Mechanical Energy = Potential Energy + Kinetic Energy

Engineers must understand both potential and kinetic energy. A simple example would be the design of a roller coaster — a project that involves both mechanical and civil engineers. At the beginning of the roller coaster, the cars must have enough potential energy to power them for the rest of the ride. This can be done by raising the cars to a great height. Then, the increased potential energy of the cars is converted into enough kinetic energy to keep them in motion for the length of the track. This is why roller coaters usually start with a big hill. As the cars start down the first hill, potential energy is changed into kinetic energy and the cars pick up speed. Engineers design the roller coaster to have enough energy to complete the course and to overcome the energy-draining effect of friction.

Watch this activity on YouTube

Restate that both potential energy and kinetic energy are forms of mechanical energy. Potential energy is the energy of position and kinetic energy is the energy of motion. A ball that you hold in your hand has potential energy, while a ball that you throw has kinetic energy. These two forms of energy can be transformed back and forth. When you drop a ball, you demonstrate an example of potential energy changing into kinetic energy.

Explain that energy is an important engineering concept. Engineers need to understand the many different forms of energy so that they can design useful products. An electric pencil sharpener serves to illustrate the point. First, the designer needs to know the amount of kinetic energy the spinning blades need in order to successfully shave off the end of the pencil. Then, the designer must choose an appropriately-powered motor to supply the necessary energy. Finally, the designer must know the electrical energy requirements of the motor in order for the motor to properly do its assigned task.

conservation of energy: A principle stating that the total energy of an isolated system remains constant regardless of changes within the system. Energy can neither be created nor destroyed.

energy: Energy is the capacity to do work.

kinetic energy: The energy of motion.

mechanical energy: Energy that is composed of both potential energy and kinetic energy.

potential energy: The energy of position, or stored energy.

Pre-Lesson Assessment

Discussion Questions: Solicit, integrate and summarize student responses.

• What are examples of dangerous unsafe placement of objects? (Possible answers: Boulders on the edge of a cliff, dishes barely on shelves, etc.).

Post-Introduction Assessment

Question/Answer: Ask the students and discuss as a class:

• What has more potential energy: a boulder on the ground or a feather 10 feet in the air? (Answer: The feather because the boulder is on the ground and has zero potential energy. However, if the boulder was 1 mm off the ground, it would probably have more potential energy.)

Lesson Summary Assessment

Group Brainstorm: Give groups of students each a ball (example, tennis ball). Remind them that energy can be converted from potential to kinetic and vice versa. Write a question on the board and have them brainstorm the answer in their groups. Have the students record their answers in their journals or on a sheet of paper and hand it in. Discuss the student groups' answers with the class.

• How can you throw a ball and have its energy change from kinetic to potential and back to kinetic without touching the ball once it relases from your hand? (Answer: Throw it straight up in the air.)

Calculating: Have students practice problems solving for potential energy and kinetic energy:

• If a mass that weighs 8 kg is held at a height of 10 m, what is its potential energy? (Answer: PE = (8 kg)*(9.8 m/s 2 )*(10 m) = 784 kg*m 2 /s 2 = 784 J)
• Now consider an object with a kinetic energy of 800 J and a mass of 12 kg. What is its velocity? (Answer: v = sqrt(2*KE/m) = sqrt((2 * 800 J)/12 kg) = 11.55 m/s)

Lesson Extension Activities

There is another form of potential energy, not related to height, which is called spring potential or elastic potential energy . In this case, energy is stored when you compress or elongate a spring. Have the students search the Internet or library for the equation of spring potential energy and explain what the variables in the equation represent. The answer is

PE spring = ½ k∙x 2

where k is the spring constant measured in N/m (Newton/meters) and x is how far the spring is compressed or stretched measured in m (meters).

This activity shows students the engineering importance of understanding the laws of mechanical energy. More specifically, it demonstrates how potential energy can be converted to kinetic energy and back again. Given a pendulum height, students calculate and predict how fast the pendulum will swing ...

This activity demonstrates how potential energy (PE) can be converted to kinetic energy (KE) and back again. Given a pendulum height, students calculate and predict how fast the pendulum will swing by understanding conservation of energy and using the equations for PE and KE.

High school students learn how engineers mathematically design roller coaster paths using the approach that a curved path can be approximated by a sequence of many short inclines. They apply basic calculus and the work-energy theorem for non-conservative forces to quantify the friction along a curve...

Students explore the physics exploited by engineers in designing today's roller coasters, including potential and kinetic energy, friction and gravity. During the associated activity, students design, build and analyze model roller coasters they make using foam tubing and marbles (as the cars).

Argonne Transportation - Laser Glazing of Rails. September 29, 2003. Argonne National Laboratory, Transportation Technology R&D Center. October 15, 2003. http://www.anl.gov/index.html

Asimov, Isaac. The History of Physics. New York: Walker & Co., 1984.

Jones, Edwin R. and Richard L. Childers. Contemporary College Physics. Reading, MA: Addison-Wesley Publishing Co., 1993.

Kahan, Peter. Science Explorer: Motion, Forces, and Energy. Upper Saddle River, NJ: Prentice Hall, 2000.

Luehmann, April. Give Me Energy. June 12, 2003. Science and Mathematics Initiative for Learning Enhancement, Illinois Institute of Technology. October 15, 2003. http://www.iit.edu/~smile/ph9407.html

Nave, C.R. HyperPhysics. 2000. Department of Physics and Astronomy, Georgia State University. October 15, 2003. hyperphysics.phy-astr.gsu.edu/hbase/hframe.html

The Atoms Family - The Mummy's Tomb – Raceways. Miami Museum of Science and Space Transit Planetarium. October 15, 2003. http://www.miamisci.org/af/sln/mummy/raceways.html

Other Related Information

Browse the NGSS Engineering-aligned Physics Curriculum hub for additional Physics and Physical Science curriculum featuring Engineering.

Contributors

Supporting program, acknowledgements.

The contents of this digital library curriculum were developed under a grant from the Fund for the Improvement of Postsecondary Education (FIPSE), U.S. Department of Education and National Science Foundation GK-12 grant no. 0338326. However, these contents do not necessarily represent the policies of the Department of Education or National Science Foundation, and you should not assume endorsement by the federal government.

Last modified: December 14, 2022

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• Classical Mechanics

How to Calculate Kinetic Energy

Last Updated: April 5, 2024 Fact Checked

This article was reviewed by Grace Imson, MA . Grace Imson is a math teacher with over 40 years of teaching experience. Grace is currently a math instructor at the City College of San Francisco and was previously in the Math Department at Saint Louis University. She has taught math at the elementary, middle, high school, and college levels. She has an MA in Education, specializing in Administration and Supervision from Saint Louis University. This article has been fact-checked, ensuring the accuracy of any cited facts and confirming the authority of its sources. This article has been viewed 1,130,003 times.

There are two basic forms of energy: potential and kinetic energy. Potential energy is the energy an object has relative to the position of another object. [1] X Research source For example, if you are at the top of a hill, you have more potential energy than if you are at the bottom of the hill. Kinetic energy is the energy an object has when it is in motion. [2] X Research source Kinetic energy can be due to vibration, rotation, or translation (movement from one place to another). [3] X Research source The kinetic energy of an object can easily be determined by an equation using the mass and velocity of that object. [4] X Research source

Understanding Kinetic Energy

• Your answer should always be stated in joules (J) , which is the standard unit of measurement for kinetic energy. It is equivalent to 1 kg * m 2 /s 2 .

• Tare the balance. Before you weigh your object, you must set it to zero. Zeroing out the scale is called taring. [6] X Research source
• Place your object in the balance. Gently, place the object on the balance and record its mass in kilograms.
• If necessary, convert grams to kilograms. For the final calculation, the mass must be in kilograms.

• Velocity is defined by the equation, displacement divided by time: V = d/t . Velocity is a vector quantity, meaning it has both a magnitude and a direction. Magnitude is the number value that quantifies the speed, while the direction is the direction in which the speed takes place during motion.
• For example, an object’s velocity can be 80 m/s or -80 m/s depending on the direction of travel.
• To calculate velocity, simply divide the distance the object traveled by the time it took to travel that distance.

Calculating Kinetic Energy

• Your answer should always be stated in joules (J), which is the standard unit of measurement for kinetic energy. It is equivalent to 1 kg * m 2 /s 2 .

• KE = 0.5 x mv 2
• KE = 0.5 x 55 x (3.87) 2

• KE = 0.5 x 55 x 14.97
• KE = 411.675 J

Using Kinetic Energy to Find Velocity or Mass

• 500 J = 0.5 x 30 x v 2
• 100 J = 0.5 x m x 5 2

• Multiply mass by 0.5: 0.5 x 30 = 15
• Divide kinetic energy by the product: 500/15 = 33.33
• Square root to find velocity: 5.77 m/s
• Square the velocity: 5 2 = 25
• Multiply by 0.5: 0.5 x 25 = 12.5
• Divide kinetic energy by product: 100/12.5 = 8 kg

Calculator, Practice Problems, and Answers

Community Q&A

You Might Also Like

• ↑ https://byjus.com/physics/potential-energy/
• ↑ https://www.khanacademy.org/science/physics/work-and-energy/work-and-energy-tutorial/a/what-is-kinetic-energy
• ↑ https://sciencing.com/sources-of-kinetic-energy-12298540.html
• ↑ https://study.com/academy/lesson/tare-weight-vs-net-weight.html
• ↑ http://www.physicsclassroom.com/class/1DKin/Lesson-1/Speed-and-Velocity
• ↑ http://www.physicsclassroom.com/class/energy/u5l1c.cfm

About This Article

To calculate kinetic energy, write out a formula where kinetic energy is equal to 0.5 times mass times velocity squared. Add in the value for the mass of the object, then the velocity with which it is moving. Solve for the unknown variable. Your answer should be stated in joules, or J. If you want to learn how to solve velocity or mass using kinetic energy, keep reading the article! Did this summary help you? Yes No

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KE = 0.5 • m • v 2 where m = mass of object

v = speed of object

Kinetic energy is a scalar quantity ; it does not have a direction. Unlike velocity , acceleration , force , and momentum , the kinetic energy of an object is completely described by magnitude alone. Like work and potential energy, the standard metric unit of measurement for kinetic energy is the Joule. As might be implied by the above equation, 1 Joule is equivalent to 1 kg*(m/s)^2.

We Would Like to Suggest ...

Check Your Understanding

Use your understanding of kinetic energy to answer the following questions. Then click the button to view the answers.

1. Determine the kinetic energy of a 625-kg roller coaster car that is moving with a speed of 18.3 m/s.

KE = (0.5) * (625 kg) * (18.3 m/s) 2

KE = 1.05 x10 5 Joules

2. If the roller coaster car in the above problem were moving with twice the speed, then what would be its new kinetic energy?

KE = 0.5*m*v 2

KE = 0.5*625 kg*(36.6 m/s) 2

KE = 4.19 x 10 5 Joules

3. Missy Diwater, the former platform diver for the Ringling Brother's Circus, had a kinetic energy of 12 000 J just prior to hitting the bucket of water. If Missy's mass is 40 kg, then what is her speed?

12 000 J = (0.5) * (40 kg) * v 2

300 J = (0.5) * v 2

600 J = v 2

v = 24.5 m/s

4. A 900-kg compact car moving at 60 mi/hr has approximately 320 000 Joules of kinetic energy. Estimate its new kinetic energy if it is moving at 30 mi/hr. (HINT: use the kinetic energy equation as a " guide to thinking .")

KE = 80 000 J

The KE is directly related to the square of the speed. If the speed is reduced by a factor of 2 (as in from 60 mi/hr to 30 mi/hr) then the KE will be reduced by a factor of 4. Thus, the new KE is (320 000 J)/4 or 80 000 J.

• Internal vs. External Forces

KINETIC ENERGY PROBLEMS WORKSHEET

Problem 1 :

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

Problem 2 :

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

Problem 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?

Problem 4 :

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

Problem 5 :

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

Problem 6 :

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

1. Solution :

K.E = (1/2)mv 2

Substitute m = 150 and v = 15.

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

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

The Kinetic Energy of the object is 16875 joules.

2. Solution :

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.

3. Solution :

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.

4. Solution :

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.

5. Solution :

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.

6. Solution :

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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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

• 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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19.4: Heat Capacity and Equipartition of Energy

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Learning Objectives

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

• Solve problems involving heat transfer to and from ideal monatomic gases whose volumes are held constant
• Solve similar problems for non-monatomic ideal gases based on the number of degrees of freedom of a molecule
• Estimate the heat capacities of metals using a model based on degrees of freedom

In the chapter on temperature and heat, we defined the specific heat capacity with the equation \(Q = mc\Delta T\), or \(c = (1/m)Q/\Delta T\). However, the properties of an ideal gas depend directly on the number of moles in a sample, so here we define specific heat capacity in terms of the number of moles, not the mass. Furthermore, when talking about solids and liquids, we ignored any changes in volume and pressure with changes in temperature—a good approximation for solids and liquids, but for gases, we have to make some condition on volume or pressure changes. Here, we focus on the heat capacity with the volume held constant. We can calculate it for an ideal gas.

Heat Capacity of an Ideal Monatomic Gas at Constant Volume

We define the molar heat capacity at constant volume \(C_V\) as

\[\underbrace{C_V = \dfrac{1}{n} \dfrac{Q}{\Delta T}}_{\text{with constant V}}\nonumber \]

This is often expressed in the form

\[Q = nC_V\Delta T\nonumber \]

If the volume does not change, there is no overall displacement, so no work is done, and the only change in internal energy is due to the heat flow \(\Delta E_{int} = Q\). (This statement is discussed further in the next chapter.) We use the equation \(E_{int} = 3nRT/2\) to write \(\Delta E_{int} = 3nR\Delta T/2\) and substitute \(\Delta E\) for Q to find \(Q = 3nR\Delta T/2\), which gives the following simple result for an ideal monatomic gas:

\[C_V = \dfrac{3}{2}R.\nonumber \]

It is independent of temperature, which justifies our use of finite differences instead of a derivative. This formula agrees well with experimental results.

In the next chapter we discuss the molar specific heat at constant pressure \(C_p\), which is always greater than \(C_V\).

Example \(\PageIndex{1}\): Calculating Temperature

A sample of 0.125 kg of xenon is contained in a rigid metal cylinder, big enough that the xenon can be modeled as an ideal gas, at a temperature of \(20.0^oC\). The cylinder is moved outside on a hot summer day. As the xenon comes into equilibrium by reaching the temperature of its surroundings, 180 J of heat are conducted to it through the cylinder walls. What is the equilibrium temperature? Ignore the expansion of the metal cylinder.

• Identify the knowns: We know the initial temperature \(T_1\) is \(20.0^oC\), the heat Q is 180 J, and the mass m of the xenon is 0.125 kg.
• Identify the unknown. We need the final temperature, so we’ll need \(\Delta T\).
• Determine which equations are needed. Because xenon gas is monatomic, we can use \(Q = 3nR\Delta T/2\). Then we need the number of moles \(n = m/M\).
• Substitute the known values into the equations and solve for the unknowns.

The molar mass of xenon is 131.3 g, so we obtain

\[n = \dfrac{125 \, g}{131.3 \, g/mol} = 0.952 \, mol, \nonumber\nonumber \]

\[\Delta T = \dfrac{2Q}{3nR} = \dfrac{2(180 \, J)}{3(0.952 \, mol)(8.31 \, J/mol \cdot \, ^oC)} = 15.2^oC. \nonumber\nonumber \]

Therefore, the final temperature is \(35.2^oC \). The problem could equally well be solved in kelvin; as a kelvin is the same size as a degree Celsius of temperature change, you would get \(\Delta T = 15.2 \, K \).

Significance

The heating of an ideal or almost ideal gas at constant volume is important in car engines and many other practical systems.

Exercise \(\PageIndex{1}\)

Suppose 2 moles of helium gas at 200 K are mixed with 2 moles of krypton gas at 400 K in a calorimeter. What is the final temperature?

As the number of moles is equal and we know the molar heat capacities of the two gases are equal, the temperature is halfway between the initial temperatures, 300 K.

We would like to generalize our results to ideal gases with more than one atom per molecule. In such systems, the molecules can have other forms of energy beside translational kinetic energy, such as rotational kinetic energy and vibrational kinetic and potential energies. We will see that a simple rule lets us determine the average energies present in these forms and solve problems in much the same way as we have for monatomic gases.

Degrees of Freedom

In the previous section, we found that \(\frac{1}{2}mv^2 = \frac{3}{2}k_BT\) and \(v^2 = 3v_x^2\), from which it follows that \(\frac{1}{2}mv_x^2 = \frac{1}{2}k_BT\). The same equation holds for \(\frac{3}{2}k_BT\) as the sum of contributions of \(\frac{1}{2}k_BT\) from each of the three dimensions of translational motion. Shifting to the gas as a whole, we see that the 3 in the formula \(C_V = \frac{3}{2}R\) also reflects those three dimensions. We define a degree of freedom as an independent possible motion of a molecule, such as each of the three dimensions of translation. Then, letting d represent the number of degrees of freedom, the molar heat capacity at constant volume of a monatomic ideal gas is \(C_V = \frac{d}{2}R\), where \(d = 3\).

The branch of physics called statistical mechanics tells us, and experiment confirms, that \(C_V\) of any ideal gas is given by this equation, regardless of the number of degrees of freedom. This fact follows from a more general result, the equipartition theorem , which holds in classical (non-quantum) thermodynamics for systems in thermal equilibrium under technical conditions that are beyond our scope. Here, we mention only that in a system, the energy is shared among the degrees of freedom by collisions.

Equipartition Theorem

The energy of a thermodynamic system in equilibrium is partitioned equally among its degrees of freedom. Accordingly, the molar heat capacity of an ideal gas is proportional to its number of degrees of freedom, d : \[C_V = \dfrac{d}{2}R.\nonumber \]

This result is due to the Scottish physicist James Clerk Maxwell (1831−1871), whose name will appear several more times in this book.

For example, consider a diatomic ideal gas (a good model for nitrogen, \(N_2\), and oxygen, \(O_2\)). Such a gas has more degrees of freedom than a monatomic gas. In addition to the three degrees of freedom for translation, it has two degrees of freedom for rotation perpendicular to its axis. Furthermore, the molecule can vibrate along its axis. This motion is often modeled by imagining a spring connecting the two atoms, and we know from simple harmonic motion that such motion has both kinetic and potential energy. Each of these forms of energy corresponds to a degree of freedom, giving two more.

We might expect that for a diatomic gas, we should use 7 as the number of degrees of freedom; classically, if the molecules of a gas had only translational kinetic energy, collisions between molecules would soon make them rotate and vibrate. However, as explained in the previous module, quantum mechanics controls which degrees of freedom are active. The result is shown in Figure \(\PageIndex{1}\). Both rotational and vibrational energies are limited to discrete values. For temperatures below about 60 K, the energies of hydrogen molecules are too low for a collision to bring the rotational state or vibrational state of a molecule from the lowest energy to the second lowest, so the only form of energy is translational kinetic energy, and \(d = 3\) or \(C_V = 3R/2\) as in a monatomic gas. Above that temperature, the two rotational degrees of freedom begin to contribute, that is, some molecules are excited to the rotational state with the second-lowest energy. (This temperature is much lower than that where rotations of monatomic gases contribute, because diatomic molecules have much higher rotational inertias and hence much lower rotational energies.) From about room temperature (a bit less than 300 K) to about 600 K, the rotational degrees of freedom are fully active, but the vibrational ones are not, and \(d = 5\). Then, finally, above about 3000 K, the vibrational degrees of freedom are fully active, and \(d = 7\) as the classical theory predicted.

Polyatomic molecules typically have one additional rotational degree of freedom at room temperature, since they have comparable moments of inertia around any axis. Thus, at room temperature, they have \(d = 6\) and at high temperature, \(d = 8\). We usually assume that gases have the theoretical room-temperature values of d .

As shown in Table \(\PageIndex{1}\), the results agree well with experiments for many monatomic and diatomic gases, but the agreement for triatomic gases is only fair. The differences arise from interactions that we have ignored between and within molecules.

What about internal energy for diatomic and polyatomic gases? For such gases, \(C_V\) is a function of temperature (Figure \(\PageIndex{1}\)), so we do not have the kind of simple result we have for monatomic ideal gases.

Molar Heat Capacity of Solid Elements

The idea of equipartition leads to an estimate of the molar heat capacity of solid elements at ordinary temperatures. We can model the atoms of a solid as attached to neighboring atoms by springs (Figure \(\PageIndex{2}\)).

Analogously to the discussion of vibration in the previous module, each atom has six degrees of freedom: one kinetic and one potential for each of the x -, y -, and z -directions. Accordingly, the molar specific heat of a metal should be 3 R . This result, known as the Law of Dulong and Petit , works fairly well experimentally at room temperature. (For every element, it fails at low temperatures for quantum-mechanical reasons. Since quantum effects are particularly important for low-mass particles, the Law of Dulong and Petit already fails at room temperature for some light elements, such as beryllium and carbon. It also fails for some heavier elements for various reasons beyond what we can cover.)

Problem-Solving Strategy: Heat Capacity and Equipartition

The strategy for solving these problems is the same as the one in Phase Changes for the effects of heat transfer. The only new feature is that you should determine whether the case just presented—ideal gases at constant volume—applies to the problem. (For solid elements, looking up the specific heat capacity is generally better than estimating it from the Law of Dulong and Petit.) In the case of an ideal gas, determine the number d of degrees of freedom from the number of atoms in the gas molecule and use it to calculate \(C_V\) (or use \(C_V\) to solve for d ).

Example \(\PageIndex{2}\): Calculating Temperature: Calorimetry with an Ideal Gas

A 300-g piece of solid gallium (a metal used in semiconductor devices) at its melting point of only \(30.0^oC\) is in contact with 12.0 moles of air (assumed diatomic) at \(95.0^oC\) in an insulated container. When the air reaches equilibrium with the gallium, 202 g of the gallium have melted. Based on those data, what is the heat of fusion of gallium? Assume the volume of the air does not change and there are no other heat transfers.

We’ll use the equation \(Q_{hot} + Q_{cold} = 0\). As some of the gallium doesn’t melt, we know the final temperature is still the melting point. Then the only \(Q_{hot}\) is the heat lost as the air cools, \(Q_{hot} = n_{air}C_V\Delta T\), where \(C_V = 5R/2\). The only \(Q_{cold}\) is the latent heat of fusion of the gallium, \(Q_{cold} = m_{Ga}L_f\). It is positive because heat flows into the gallium.

• Set up the equation:\[n_{air}C_V\Delta T + m_{Ga}L_f = 0.\nonumber \]
• Substitute the known values and solve: \[(12.0 \, mol) \left(\dfrac{5}{2}\right) \left(8.31 \dfrac{J}{mol \cdot \, ^oC}\right)(30.0^oC - 95.0^oC) + (0.202 \, kg)L_f = 0.\nonumber \]

We solve to find that the heat of fusion of gallium is 80.2 kJ/kg.

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Solving wind’s dirty secret: innovating wind turbine blade disposal

With 800,000 tonnes (t) of turbine blades disposed of in landfill annually, the wind industry has an environmental problem to solve.

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The disposal of wind turbine blades that have come to the end of their working lives is posing an environmental problem for an industry that is intended to help mitigate such impacts.

Made from fibreglass-reinforced polymer (FRP) and coated with epoxy resins, turbine blades are designed to maximise aerodynamicism while remaining light enough to minimise structural stress and stiff enough to achieve efficient wind capture.

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The problem is, having been designed to withstand extreme weather conditions, epoxy resin is durable and therefore difficult to breakdown. The tough coating makes it near-impossible to separate the materials comprising a blade. Separating the fibres from the polymer using high temperatures results in the degradation of the plastic parts and damage to the glass fibres.

More than 317,000 onshore wind turbines are active globally, while an additional 12,000 are active offshore, according to figures from Energy Monitor ‘s parent company GlobalData. Since the first wind farm was erected in the US state of New Hampshire in 1980, more than 12,600 turbines have so far been decommissioned. However, that number is set to grow rapidly , as many are nearing the end of their 20–30-year lifespan.

The EU’s Waste Framework Directive specifies that landfill is the “least preferred waste management option”, calling for prevention and preparation for reuse, recycling and recovery. With an estimated 800,000 tonnes of turbine blades entering landfill annually , however, there is a serious sustainability challenge to overcome.

Continuum: a mechanical solution

Although a problem for the industry, the issue of sustainable disposal of composite wind turbine blade waste has presented an opportunity for innovators. This includes Continuum, which works with machinery partner Dieffenbacher to convert fibreglass waste into construction materials.

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Explaining the company’s approach to Power Technology , Martin Dronfield, CEO and interim chair of Continuum, says: “Using a 100% mechanical recycling process based around a complex series of machines”, the ‘front end’ of the process “mechanically strips back the end-of-life composite to small particles and constituent parts”, while the ‘back end’ “reconstructs the material with a new virgin resin and transforms it into high-performance, high-value construction panels used for building facades, concrete formwork, doors and floors and wet room wall materials”.

He adds: “The panel has an CO₂ footprint of less than 70kg/t, while the zero CO₂-emissions factory takes in 36,000t of end-of-life thermoset composite annually, and produces 30,000m³ of high-performance panelling, again annually.”

The company hopes to operate six factories across Europe by 2030, producing infinitely recyclable panels made of 92% recycled blades, with the other 8% comprised of resin. The process is powered by green electricity and produces no emissions, no waste liquids and emits no dust.

Vestas Wind Systems: a chemical solution

Vestas Wind Systems has opted for a different technique, announcing a “newly discovered chemical process” in February 2023. The method breaks down the blades to produce high-quality materials, which can be repurposed to make new blades, potentially offering circularity to the industry.

In the announcement, Lisa Ekstrand, the company’s vice-president and head of sustainability, said: “Once this new technology is implemented at scale, legacy blade material currently sitting in landfill, as well as blade material in active windfarms, can be disassembled and reused. This signals a new era for the wind industry, and accelerates our journey towards achieving circularity.”

However, in a review of recycling technologies for FRP conducted in 2023, researchers pointed out that while chemical processes “are more prone to industrial scalability thanks to the high volume of recycling material processed”, there remains an obstacle – namely “high energy consumption”.

Instead, the review favours mechanical processes for managing waste – such as those used by Continuum – noting that “among the recycling methods treated in this review, the mechanical ones, such as shredding, hammer milling, milling and grinding, are the more appreciable recycling processes to recover glass fibres”.

LM Wind Power: a preventative solution

Considering leaders in turbine blade disposal, GlobaData practice head Mohit Prasad highlights LM Wind Power as a company that “will play a central role in supporting their customers to develop fully circular wind turbines that generate less waste during their production”.

The company addresses waste at the other end of a blade’s lifespan, aiming to eliminate waste during the manufacturing process. It reports that 20–25% of materials purchased by wind turbine blade manufacturers are not used in the final product and notes: “Research indicates that blade manufacturing waste volumes are expected to be larger than decommissioned blade volumes in the coming decade.”

Indeed, previous estimates have found that “waste generated during manufacturing and service adds between 16% and 45% of the mass of the wind turbine blades”.

LM Wind Power is attempting to reduce this by manufacturing zero-waste blades by 2030. It promises to create a circular economy by eliminating all waste – usually comprising 30% process materials (consumables, mixed with resin), 30% packaging, 15% dry glass fibre, 10% cut offs to composite materials and 15% other, it reports – from the manufacturing process and from packaging.

Polishing a clean industry

GlobalData’s Prasad tells Power Technology that the wind turbine industry is already fairly clean, as 85–95% of turbine’s materials are recycled. He explains that: “Masts, cables, gearboxes and generators are recycled and recovered,” while “foundations which are made of concrete are crushed to be used as filler”.

Nacelles and blades are generally made of fibreglass and are “ground down to be used as fuel in cement factories or filler in road construction”. Prasad notes that innovative solutions such as repurposing blades into playgrounds or bike sheds, “have been shown to be effective at a local level”.

However, local solutions are rapidly becoming inadequate. According to the Union of Concerned Scientists, the average blade size for an onshore wind turbine is around 164ft in length , while offshore turbines tend to be larger, with blade spans of up to 260–290ft in length . Repurposing these is not straightforward.

Yet, despite their massive size and landfill requirements, turbine blades are responsible for providing between 120,000 and 180,000 hours of electricity generation in their lifespans, making them a crucial part of the energy transition, according to Prasad. Innovation at all stages of the life cycle by companies such as Continuum, Vestas Wind Systems and LM Wind Power promises to clean up the sector’s waste challenge.

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