Contents

Kinetic Energy and 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. $F_{net} F_{net} size 12{F rSub { size 8{"net"} } } {}$ . In equation form, this is $W_{net} = F_{net} d cos θ W_{net} = F_{net} d size 12{W rSub { size 8{"net"} } =F rSub { size 8{"net"} } d"cos"θ} {}$ where $θ θ size 12{θ} {}$ 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 $F cos θ F size 12{F"cos"θ} {}$ vs. $d d size 12{d} {}$ graph. In this case, $F cos θ F size 12{F"cos"θ} {}$ is constant. You can see that the area under the graph is $F d cos θ F d size 12{F"cos"θ} {}$ , 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 $(F cos θ)_{i (ave)} (F size 12{ $ F"cos"θ $ rSub { size 8{i $ "ave" $ } } } {}$ . The work done is $(F cos θ)_{i (ave)} d_{i} (F 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 $W_{i} W_{i} size 12{W rSub { size 8{i} } } {}$ . Thus the total work done is the total area under the curve, a useful property to which we shall refer later.

Two drawings labele a and b. (a) A graph of force component F cosine theta versus distance d. d is along the x axis and F cosine theta is along the y axis. A line of length d is drawn parallel to the horizontal axis for some value of F cosine theta. Area under this line in the graph is shaded and is equal to F cosine theta multiplied by d. F d cosine theta is equal to work W. (b) A graph of force component F cosine theta versus distance d. d is along the x axis and F cosine theta is along the y axis. There is an inclined line and the area under it is divided into many thin vertical strips of width d sub i. The area of one vertical stripe is equal to average value of F cosine theta times d sub i which equals to work W sub i.

Figure 7.3 (a) A graph of

F cos θ F

vs.

d d size 12{d} {}

, when

F cos θ F size 12{F"cos"θ} {}

is constant. The area under the curve represents the work done by the force. (b) A graph of

F cos θ F size 12{F"cos"q} {}

vs.

d d 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.

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.

A package shown on a roller belt pushed with a force F towards the right shown by a vector F sub app equal to one hundred and twenty newtons. A vector w is in the downward direction starting from the bottom of the package and the reaction force N on the package is shown by the vector N pointing upwards at the bottom of the package. A frictional force vector of five point zero zero newtons acts on the package leftwards. The displacement d is shown by the vector pointing to the right with a value of zero point eight zero zero meters.

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 $F_{app} F_{app}$ 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 $cos θ = 1 size 12{"cos"q=1} {}$ , and the net work is given by

W_{net} = F_{net} d . W_{net} = F_{net} d . size 12{W rSub { size 8{"net"} } =F rSub { size 8{"net"} } d} {}

7.7

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

W_{net} = mad. W_{net} = 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 $d = x - x_{0} d = x - x_{0} size 12{d=x - x rSub { size 8{0} } } {}$ and use the equation studied in Motion Equations for Constant Acceleration in One Dimension for the change in speed over a distance $d d$ 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 $a = \frac{v^{2} - {v_{0}}^{2}}{2 d} a = \frac{v^{2} - {v_{0}}^{2}}{2 d}$ . When $a a$ is substituted into the preceding expression for $W_{net} W_{net}$ , we obtain

W_{net} = m (\frac{v^{2} - {v_{0}}^{2}}{2 d}) d . W_{net} = m (\frac{v^{2} - {v_{0}}^{2}}{2 d}) d .

7.9

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

W = \frac{1}{2} {mv}^{2} - \frac{1}{2} {mv}_{0}^{2} . W = \frac{1}{2} {mv}^{2} - \frac{1}{2} {mv}_{0}^{2} . size 12{w"" lSub { size 8{ ital "net"} }  =  {  {1}  over  {2} }   ital "mv" rSup { size 8{2} }   -   {  {1}  over  {2} }   ital "mv""" lSub { size 8{0} } "" lSup { size 8{2} }  "." } {}

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 $\frac{1}{2} {mv}^{2} \frac{1}{2} {mv}^{2} size 12{ { {1} over {2} } ital "mv" rSup { size 8{2} } } {}$ . 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 $\frac{1}{2} {mv}^{2} \frac{1}{2} {mv}^{2} size 12{ { { size 8{1} } over { size 8{2} } } ital "mv" rSup { size 8{2} } } {}$ .

W_{net} = \frac{1}{2} {mv}^{2} - \frac{1}{2} {mv}_{0}^{2} W_{net} = \frac{1}{2} {mv}^{2} - \frac{1}{2} {mv}_{0}^{2} size 12{w"" lSub { size 8{ ital "net"} }  =  {  {1}  over  {2} }   ital "mv" rSup { size 8{2} }   -   {  {1}  over  {2} }   ital "mv""" lSub { size 8{0} } "" lSup { size 8{2} }  "." } {}

7.11

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

KE = \frac{1}{2} {mv}^{2}, KE = \frac{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.

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 $m m$ and speed $v v$ are given, the kinetic energy can be calculated from its definition as given in the equation $KE = \frac{1}{2} {mv}^{2} KE = \frac{1}{2} {mv}^{2} size 12{"KE"= { {1} over {2} } ital "mv" rSup { size 8{2} } } {}$ .

Solution

The kinetic energy is given by

KE = \frac{1}{2} {mv}^{2} . KE = \frac{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 \cdot m^{2} {/s}^{2} = 3.75 J. KE = 3.75 kg \cdot 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.

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

\begin{array}{l} W_{net} & = & F_{net} d = (115 N) (0.800 m) \\ = & 92.0 N \cdot m = 92.0 J. \end{array} \begin{array}{l} W_{net} & = & F_{net} d = (115 N) (0.800 m) \\ = 92.0 N \cdot m = 92.0 J. \end{array} 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" "." 0`N cdot m="92" "." 0`J "."  {}
} } {}

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.

\begin{array}{l} W_{app} & = & F_{app} d cos (0º) = F_{app} d \\ = & (120 N) (0.800 m) \\ = & 96.0 J \end{array} \begin{array}{l} W_{app} & = & F_{app} d \\ = (120 N) (0.800 m) \\ = 96.0 J \end{array} 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 $θ = 180º θ = 180º size 12{θ="180"°} {}$ , and the work done by friction is

\begin{array}{l} W_{fr} & = & F_{fr} d cos (180º) = - F_{fr} d \\ = & - (5.00 N) (0.800 m) \\ = & - 4.00 J. \end{array} \begin{array}{l} W_{fr} & = & F_{fr} d \\ = - (5.00 N) (0.800 m) \\ = - 4.00 J. \end{array} 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,

\begin{array}{l} W_{gr} & = & 0, \\ W_{N} & = & 0, \\ W_{app} & = & 96.0 J, \\ W_{fr} & = & - 4.00 J. \end{array} \begin{array}{l} W_{gr} & = & 0, \\ W_{N} & = & 0, \\ W_{app} & = & 96.0 J, \\ W_{fr} & = & - 4.00 J. \end{array} 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

W_{total} = W_{gr} + W_{N} + W_{app} + W_{fr} = 92 .0 J . W_{total} = W_{gr} + W_{N} + W_{app} + W_{fr} = 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 $W_{total} W_{total} size 12{W rSub { size 8{"total"} } } {}$ as the sum of the work by each force agrees, as expected, with the work $W_{net} W_{net} size 12{W rSub { size 8{"net"} } } {}$ 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, $W_{net} W_{net} size 12{W rSub { size 8{"net"} } } {}$ , and the initial kinetic energy, $\frac{1}{2} {m v_{0}}^{2} \frac{1}{2} {m v_{0}}^{2} size 12{ { {1} over {2} } ital "mv" rSub { size 8{0} rSup { size 8{2} } } } {}$ . These calculations allow us to find the final kinetic energy, $\frac{1}{2} {mv}^{2} \frac{1}{2} {mv}^{2} size 12{ { {1} over {2} } ital "mv" rSup { size 8{2} } } {}$ , and thus the final speed $v v size 12{v} {}$ .

Solution

The work-energy theorem in equation form is

W_{net} = \frac{1}{2} {mv}^{2} - \frac{1}{2} {m v_{0}}^{2} . W_{net} = \frac{1}{2} {mv}^{2} - \frac{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 $\frac{1}{2} {mv}^{2} \frac{1}{2} {mv}^{2} size 12{ { {1} over {2} } ital "mv" rSup { size 8{2} } } {}$ gives

\frac{1}{2} {mv}^{2} = W_{net} + \frac{1}{2} {m v_{0}}^{2} . \frac{1}{2} {mv}^{2} = W_{net} + \frac{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,

\frac{1}{2} {mv}^{2} = 92 . 0 J + 3.75 J= 95.75 J. \frac{1}{2} {mv}^{2} = 92 . 0 J + 3.75 J= 95.75 J. size 12{ {  {1}  over  {2} }  ital "mv" rSup { size 8{2} } ="92" "." 0`J+3 "." "75"`J="95" "." "75"`J} {}

7.23

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

\begin{array}{l} v & = & \sqrt{\frac{2 (95.75 J)}{m}} = \sqrt{\frac{191.5 kg \cdot m^{2} {/s}^{2}}{30.0 kg}} \\ = & 2.53 m/s. \end{array} \begin{array}{l} v & = & \sqrt{\frac{2 (95.75 J)}{m}} = \sqrt{\frac{191.5 kg \cdot m^{2} {/s}^{2}}{30.0 kg}} \\ = 2.53 m/s. \end{array}

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 $θ = 180º θ = 180º$ . 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, $W_{fr} = f d' cos θ = – f d' W_{fr} = f d'$ , where $d' d'$ is the distance it takes to stop. Thus,

d' = - \frac{W_{fr}}{f} = - \frac{- 95.75 J}{5.00 N}, d' = - \frac{W_{fr}}{f} = - \frac{- 95.75 J}{5.00 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.