 College Physics for AP® Courses

# 7.2Kinetic Energy and the Work-Energy Theorem

College Physics for AP® Courses7.2 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.

The information presented in this section supports the following AP® learning objectives and science practices:

• 3.E.1.1 The student is able to make predictions about the changes in kinetic energy of an object based on considerations of the direction of the net force on the object as the object moves. (S.P. 6.4, 7.2)
• 3.E.1.2 The student is able to use net force and velocity vectors to determine qualitatively whether kinetic energy of an object would increase, decrease, or remain unchanged. (S.P. 1.4)
• 3.E.1.3 The student is able to use force and velocity vectors to determine qualitatively or quantitatively the net force exerted on an object and qualitatively whether kinetic energy of that object would increase, decrease, or remain unchanged. (S.P. 1.4, 2.2)
• 3.E.1.4 The student is able to apply mathematical routines to determine the change in kinetic energy of an object given the forces on the object and the displacement of the object. (S.P. 2.2)
• 4.C.1.1 The student is able to calculate the total energy of a system and justify the mathematical routines used in the calculation of component types of energy within the system whose sum is the total energy. (S.P. 1.4, 2.1, 2.2)
• 4.C.2.1 The student is able to make predictions about the changes in the mechanical energy of a system when a component of an external force acts parallel or antiparallel to the direction of the displacement of the center of mass. (S.P. 6.4)
• 4.C.2.2 The student is able to apply the concepts of conservation of energy and the work-energy theorem to determine qualitatively and/or quantitatively that work done on a two-object system in linear motion will change the kinetic energy of the center of mass of the system, the potential energy of the systems, and/or the internal energy of the system. (S.P. 1.4, 2.2, 7.2)
• 5.B.5.3 The student is able to predict and calculate from graphical data the energy transfer to or work done on an object or system from information about a force exerted on the object or system through a distance. (S.P. 1.5, 2.2, 6.4)

### 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 done on an object. The net work can be written in terms of the net force on an object. $FnetFnet size 12{F rSub { size 8{"net"} } } {}$ In equation form, this is $Wnet=FnetdcosθWnet=Fnetdcosθ size 12{W rSub { size 8{"net"} } =F rSub { size 8{"net"} } d"cos"θ} {}$ where $θθ size 12{θ} {}$ 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 $FcosθFcosθ size 12{F"cos"θ} {}$ vs. $dd size 12{d} {}$ graph. In this case, $FcosθFcosθ size 12{F"cos"θ} {}$ is constant. You can see that the area under the graph is $FdcosθFdcosθ size 12{F"cos"θ} {}$, 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 $(Fcosθ)i(ave)(Fcosθ)i(ave) size 12{ $$F"cos"θ$$ rSub { size 8{i $$"ave"$$ } } } {}$. The work done is $(Fcosθ)i(ave)di(Fcosθ)i(ave)di size 12{ $$F"cos"θ$$ rSub { size 8{i $$"ave"$$ } } d rSub { size 8{i} } } {}$ for each strip, and the total work done is the sum of the $WiWi size 12{W rSub { size 8{i} } } {}$. Thus the total work done is the total area under the curve, a useful property to which we shall refer later.

Figure 7.3 (a) A graph of $FcosθFcosθ$ vs. $dd size 12{d} {}$, when $FcosθFcosθ size 12{F"cos"θ} {}$ is constant. The area under the curve represents the work done by the force. (b) A graph of $FcosθFcosθ size 12{F"cos"q} {}$ vs. $dd size 12{d} {}$ in which the force varies. The work done for each interval is the area of each strip; thus, the total area under the curve equals the total work done.

### Real World Connections: Work and Direction

Consider driving in a car. While moving, you have forward velocity and therefore kinetic energy. When you hit the brakes, they exert a force opposite to your direction of motion (acting through the wheels). The brakes do work on your car and reduce the kinetic energy. Similarly, when you accelerate, the engine (acting through the wheels) exerts a force in the direction of motion. The engine does work on your car, and increases the kinetic energy. Finally, if you go around a corner at a constant speed, you have the same kinetic energy both before and after the corner. The force exerted by the engine was perpendicular to the direction of motion, and therefore did no work and did not change the kinetic energy.

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.

Figure 7.4 A package on a roller belt is pushed horizontally through a distance $d d$.

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 $FappFapp$ and the horizontal friction force $ff$. Thus, as expected, the net force is parallel to the displacement, so that $θ=0ºθ=0º$ and $cosθ=1cosθ=1 size 12{"cos"q=1} {}$, and the net work is given by

$Wnet=Fnetd.Wnet=Fnetd. size 12{W rSub { size 8{"net"} } =F rSub { size 8{"net"} } d} {}$
7.7

The effect of the net force $FnetFnet size 12{F rSub { size 8{"net"} } } {}$ is to accelerate the package from $v0v0 size 12{v rSub { size 8{0} } } {}$ to $vv size 12{v} {}$. 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 $Fnet=maFnet=ma size 12{F rSub { size 8{"net"} } = ital "ma"} {}$ from Newton’s second law gives

$Wnet=mad.Wnet=mad. size 12{W rSub { size 8{"net"} } = ital "mad"} {}$
7.8

To get a relationship between net work and the speed given to a system by the net force acting on it, we take $d=x−x0d=x−x0 size 12{d=x - x rSub { size 8{0} } } {}$ and use the equation studied in Motion Equations for Constant Acceleration in One Dimension for the change in speed over a distance $dd$ if the acceleration has the constant value $aa$; namely, $v2=v02+2adv2=v02+2ad$ (note that $aa$ appears in the expression for the net work). Solving for acceleration gives $a=v2−v022da=v2−v022d$. When $aa$ is substituted into the preceding expression for $WnetWnet$, we obtain

$Wnet=mv2−v022dd.Wnet=mv2−v022dd.$
7.9

The $dd size 12{d} {}$ cancels, and we rearrange this to obtain

7.10

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 $12mv212mv2 size 12{ { {1} over {2} } ital "mv" rSup { size 8{2} } } {}$. 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 $12mv212mv2 size 12{ { { size 8{1} } over { size 8{2} } } ital "mv" rSup { size 8{2} } } {}$.

7.11

The quantity $12mv212mv2 size 12{ { {1} over {2} } ital "mv" rSup { size 8{2} } } {}$ in the work-energy theorem is defined to be the translational kinetic energy (KE) of a mass $mm size 12{m} {}$ moving at a speed $vv size 12{v} {}$. (Translational kinetic energy is distinct from rotational kinetic energy, which is considered later.) In equation form, the translational kinetic energy,

$KE = 1 2 mv 2 , KE = 1 2 mv 2 , size 12{"KE"= { {1} over {2} } ital "mv" rSup { size 8{2} } ,} {}$
7.12

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.

### Applying the Science Practices: Cars on a Hill

Assemble a ramp suitable for rolling some toy cars up or down. Then plan a series of experiments to determine how the direction of a force relative to the velocity of an object alters the kinetic energy of the object. Note that gravity will be pointing down in all cases. What happens if you start the car at the top? How about at the bottom, with an initial velocity that is increasing? If your ramp is wide enough, what happens if you send the toy car straight across? Does varying the surface of the ramp change your results?

Sample Response: When the toy car is going down the ramp, with a component of gravity in the same direction, the kinetic energy increases. Sending the car up the ramp decreases the kinetic energy, as gravity is opposing the motion. Sending the car sideways should result in little to no change. If you have a surface that generates more friction than a smooth surface (carpet), note that the friction always opposed the motion, and hence decreases the kinetic 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?

#### Strategy

Because the mass $mm$ and speed $vv$ are given, the kinetic energy can be calculated from its definition as given in the equation $KE=12mv2KE=12mv2 size 12{"KE"= { {1} over {2} } ital "mv" rSup { size 8{2} } } {}$.

#### Solution

The kinetic energy is given by

$KE = 1 2 mv 2 . KE = 1 2 mv 2 . size 12{"KE"= { {1} over {2} } ital "mv" rSup { size 8{2} } "." } {}$
7.13

Entering known values gives

$KE = 0 . 5 ( 30.0 kg ) ( 0.500 m/s ) 2 , KE = 0 . 5 ( 30.0 kg ) ( 0.500 m/s ) 2 , size 12{"KE"=0 "." 5 $$"30" "." 0" kg"$$ $$0 "." "500"" m/s"$$ rSup { size 8{2} } ,} {}$
7.14

which yields

$KE = 3.75 kg ⋅ m 2 /s 2 = 3.75 J. KE = 3.75 kg ⋅ m 2 /s 2 = 3.75 J. size 12{"KE"=3 "." "75""kg" cdot m rSup { size 8{2} } "/s" rSup { size 8{2} } =3 "." "75"J "." } {}$
7.15

#### Discussion

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.

### Real World Connections: Center of Mass

Suppose we have two experimental carts, of equal mass, latched together on a track with a compressed spring between them. When the latch is released, the spring does 10 J of work on the carts (we’ll see how in a couple of sections). The carts move relative to the spring, which is the center of mass of the system. However, the center of mass stays fixed. How can we consider the kinetic energy of this system?

By the work-energy theorem, the work done by the spring on the carts must turn into kinetic energy. So this system has 10 J of kinetic energy. The total kinetic energy of the system is the kinetic energy of the center of mass of the system relative to the fixed origin plus the kinetic energy of each cart relative to the center of mass. We know that the center of mass relative to the fixed origin does not move, and therefore all of the kinetic energy must be distributed among the carts relative to the center of mass. Since the carts have equal mass, they each receive an equal amount of kinetic energy, so each cart has 5.0 J of kinetic energy.

In our example, the forces between the spring and each cart are internal to the system. According to Newton’s third law, these internal forces will cancel since they are equal and opposite in direction. However, this does not imply that these internal forces will not do work. Thus, the change in kinetic energy of the system is caused by work done by the force of the spring, and results in the motion of the two carts relative to the center of mass.

### 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 $Fnet = 120 N – 5.00 N = 115 NFnet = 120 N – 5.00 N = 115 N size 12{F rSub { size 8{"net"} } " = 120 N – 5" "." "00 N = 115 N"} {}$. Thus the net work is

W net = F net d = 115 N 0.800 m = 92.0 N ⋅ m = 92.0 J. W net = F net d = 115 N 0.800 m = 92.0 N ⋅ m = 92.0 J. alignl { stack { size 12{W rSub { size 8{"net"} } =F rSub { size 8{"net"} } d= left ("115"N right ) left (0 "." "800"m right )} {} # " "="92" "." 0N cdot m="92" "." 0J "." {} } } {}
7.16

#### 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.

W app = F app d cos 0º = F app d = 120 N 0.800 m = 96.0 J W app = F app d cos 0º = F app d = 120 N 0.800 m = 96.0 J alignl { stack { size 12{W rSub { size 8{"app"} } =F rSub { size 8{"app"} } d"cos" left (0° right )=F rSub { size 8{"app"} } d} {} # " "= left ("120 N" right ) left (0 "." "800"" m" right ) {} # " "=" 96" "." "0 J" "." {} } } {}
7.17

The friction force and displacement are in opposite directions, so that $θ=180ºθ=180º size 12{θ="180"°} {}$, and the work done by friction is

W fr = F fr d cos 180º = − F fr d = − 5.00 N 0.800 m = − 4.00 J. W fr = F fr d cos 180º = − F fr d = − 5.00 N 0.800 m = − 4.00 J. alignl { stack { size 12{W rSub { size 8{"fr"} } =F rSub { size 8{"fr"} } d"cos" left ("180"° right )= - F rSub { size 8{"fr"} } d} {} # " "= - left (5 "." "00 N" right ) left (0 "." "800"" m" right ) {} # ital " "= - 4 "." "00" J "." {} } } {}
7.18

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

W gr = 0, W N = 0, W app = 96.0 J, W fr = − 4.00 J. W gr = 0, W N = 0, W app = 96.0 J, W fr = − 4.00 J. alignl { stack { size 12{W rSub { size 8{"gr"} } =0,} {} # W rSub { size 8{N} } =0, {} # W rSub { size 8{"app"} } ="96" "." 0" J," {} # W rSub { size 8{"fr"} } = - 4 "." "00"" J" "." {} } } {}
7.19

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

$Wtotal=Wgr+WN+Wapp+Wfr=92.0 J.Wtotal=Wgr+WN+Wapp+Wfr=92.0 J. size 12{W rSub { size 8{"total"} } =W rSub { size 8{"gr"} } +W rSub { size 8{N} } +W rSub { size 8{"app"} } +W rSub { size 8{"fr"} } ="92" "." 0" J"} {}$
7.20

#### Discussion for (b)

The calculated total work $WtotalWtotal size 12{W rSub { size 8{"total"} } } {}$ as the sum of the work by each force agrees, as expected, with the work $WnetWnet size 12{W rSub { size 8{"net"} } } {}$ 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.

#### Strategy

Here the work-energy theorem can be used, because we have just calculated the net work, $WnetWnet size 12{W rSub { size 8{"net"} } } {}$, and the initial kinetic energy, $1 2 m v 0 2 1 2 m v 0 2 size 12{ { {1} over {2} } ital "mv" rSub { size 8{0} rSup { size 8{2} } } } {}$. These calculations allow us to find the final kinetic energy, $12mv212mv2 size 12{ { {1} over {2} } ital "mv" rSup { size 8{2} } } {}$, and thus the final speed $vv size 12{v} {}$.

#### Solution

The work-energy theorem in equation form is

$W net = 1 2 mv 2 − 1 2 m v 0 2 . W net = 1 2 mv 2 − 1 2 m v 0 2 . size 12{W rSub { size 8{"net"} } = { {1} over {2} } ital "mv" rSup { size 8{2} } - { {1} over {2} } ital "mv" rSub { size 8{0} rSup { size 8{2} } } "." } {}$
7.21

Solving for $12mv212mv2 size 12{ { {1} over {2} } ital "mv" rSup { size 8{2} } } {}$ gives

$1 2 mv 2 = W net + 1 2 m v 0 2 . 1 2 mv 2 = W net + 1 2 m v 0 2 . size 12{ { {1} over {2} } ital "mv""" lSup { size 8{2} } =w rSub { size 8{ ital "net"} } + { {1} over {2} } ital "mv""" lSub { size 8{0} } "" lSup { size 8{2} } "." } {}$
7.22

Thus,

$12mv2=92.0 J+3.75 J=95.75 J.12mv2=92.0 J+3.75 J=95.75 J. size 12{ { {1} over {2} } ital "mv" rSup { size 8{2} } ="92" "." 0J+3 "." "75"J="95" "." "75"J} {}$
7.23

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

$v = 2 (95.75 J) m = 191.5 kg ⋅ m 2 /s 2 30.0 kg = 2.53 m/s. v = 2 (95.75 J) m = 191.5 kg ⋅ m 2 /s 2 30.0 kg = 2.53 m/s.$
7.24

#### Discussion

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.

#### Strategy

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.

#### Solution

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 $θ=180ºθ=180º$. To reduce the kinetic energy of the package to zero, the work $WfrWfr$ by friction must be minus the kinetic energy that the package started with plus what the package accumulated due to the pushing. Thus $Wfr=−95.75 JWfr=−95.75 J$. Furthermore, $Wfr=fd′cosθ= –fd′Wfr=fd′cosθ= –fd′$, where $d′d′$ is the distance it takes to stop. Thus,

$d′=−Wfrf=−−95.75 J5.00 N,d′=−Wfrf=−−95.75 J5.00 N, size 12{ { {d}} sup { ' }= - { {W rSub { size 8{"fr"} } } over {f} } = - { { - "95" "." "75"J} over {5 "." "00 N"} } } {}$
7.25

and so

$d′=19.2 m.d′=19.2 m. size 12{ { {d}} sup { ' }="19" "." 2" m"} {}$
7.26

#### Discussion

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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