Chapter 9 Work and Kinetic Energy 9.1 The Concept of Energy The transformation of energy is a powerful concept that enables us to describe a vast number of processes:

• Falling water releases stored gravitational potential energy, which can become the kinetic energy associated with a coherent motion of matter. The harnessed mechanical energy can be used to spin turbines and alternators, doing work to generate electrical energy, transmitted to consumers along power lines. When you use any electrical device, the electrical energy is transformed into other forms of energy. In a refrigerator, electrical energy is used to compress a gas into a liquid. During the compression, some of the internal energy of the gas is transferred to the random motion of molecules in the outside environment. The liquid flows from a high-pressure region into a low-pressure region where the liquid evaporates. During the evaporation, the liquid absorbs energy from the random motion of molecules inside of the refrigerator. The gas returns to the compressor.

• “Human beings transform the stored chemical energy of food into various forms

necessary for the maintenance of the functions of the various organ system, tissues and cells in the body.”

1 A person can do work on their surroundings – for

example, by pedaling a bicycle – and transfer energy to the surroundings in the form of increasing random motion of air molecules, by using this catabolic energy.

• Burning gasoline in car engines converts chemical energy, stored in the molecular

bonds of the constituent molecules of gasoline, into coherent (ordered) motion of the molecules that constitute a piston. With the use of gearing and tire/road friction, this motion is converted into kinetic energy of the car; the automobile moves.

• Stretching or compressing a spring stores elastic potential energy that can be

released as kinetic energy.

• The process of vision begins with stored atomic energy released as electromagnetic radiation (light), which is detected by exciting photoreceptors in the eye, releasing chemical energy.

• When a proton fuses with deuterium (a hydrogen atom with a neutron and proton

for a nucleus), helium-three is formed (with a nucleus of two protons and one

1 George B. Benedek and Felix M.H. Villars, Physics with Illustrative Examples from Medicine and Biology, Volume 1: Mechanics, Addison-Wesley, Reading, 1973, p. 5-116

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neutron) along with radiant energy in the form of photons. The combined internal energy of the proton and deuterium are greater than the internal energy of the helium-three. This difference in internal energy is carried away by the photons as light energy.

There are many such processes in the manmade and natural worlds, involving different forms of energy: kinetic energy, gravitational energy, heat energy, elastic energy, electrical energy, chemical energy, electromagnetic energy, nuclear energy and more. The total energy is always conserved in these processes, although different forms of energy are converted into others. Any physical process can be characterized by two states, initial and final, between which energy transformations can occur. Each form of energy , where “ ” is an arbitrary label identifying one of the forms of energy, may undergo a change during this transformation,

iE iN

final, initial,i iE E E iΔ ≡ − . (9.1.1) Conservation of energy means that the sum of these changes is zero,

. (9.1.2) 1 21

0N

N ii

E E E E=

Δ + Δ + ⋅⋅⋅ + Δ = Δ =∑ Two important points emerge from this idea. First, we are interested primarily in changes in energy and so we search for relations that describe how each form of energy changes. Second, we must account for all the ways energy can change. If we observe a process, and the sum of the changes in energy is not zero, either our expressions for energy are incorrect, or there is a new type of change of energy that we had not previously discovered. This is our first example of the importance of conservation laws in describing physical processes, as energy is a key quantity conserved in all physical processes. If we can quantify the changes of different forms of energy, we have a very powerful tool to understand nature. We will begin our analysis of conservation of energy by considering processes involving only a few forms of changing energy. We will make assumptions that greatly simplify our description of these processes. At first we shall only consider processes acting on bodies in which the atoms move in a coherent fashion, ignoring processes in which energy is transferred into the random motion of atoms. Thus we will initially ignore the effects of friction. We shall then treat processes involving friction between rigid bodies. We will later return to processes in which there is an energy transfer resulting in an increase or decrease in random motion when we study the First Law of Thermodynamics.

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Energy is always conserved but we often prefer to restrict our attention to a set of objects that we define to be our system. The rest of the universe acts as the surroundings. We illustrate this division of system and surroundings in Figure 9.1.

Figure 9.1 A diagram of a system and its surroundings with boundary Since energy is conserved, any energy that leaves the system must cross through the boundary and enter the surroundings. Consider any physical process that occurs between an initial state and a final state, in which energy transformations occur. Our statement of conservation of energy becomes , (9.1.3) total system surroundings 0E E EΔ = Δ + Δ = where (9.1.4) system final, system initial, systemE E EΔ ≡ − is the change in energy of the system and (9.1.5) surroundings final, surroundings initial, surroundingsE E EΔ ≡ − is the change in energy of the surroundings. 9.2 Kinetic Energy The first form of energy that we will study is an energy associated with the coherent motion of molecules that constitute a body of mass ; this energy is called the kinetic energy (from the Greek “kinetikos,” moving). Let us consider a car moving along a straight road (along which we will place the

m

x -axis). For an observer at rest with respect to the ground, the car has velocity ˆ

xv=v i . The speed of the car is the magnitude

of the velocity, xv v= . Definition: Kinetic Energy

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The kinetic energy of a non-rotating body of mass moving with speed is defined to be the positive scalar quantity

K m v

212

K mv≡ . (9.2.1)

The kinetic energy is proportional to the square of the speed. The SI units for kinetic energy are . This combination of units is defined to be a

joule and is denoted by [J], thus

2 2kg m s−⎡ ⋅ ⋅⎣ ⎤⎦221 J 1 kg m s−≡ ⋅ ⋅ . (The SI unit of energy is

named for James Prescott Joule; the proper pronunciation of the unit “joule” is “jowl”, but the far more common usage is the incorrect “jool”.)

The above definition of kinetic energy does not refer to any direction of motion, just the speed of the body. We will see that it is sometimes convenient to express (9.2.1) in the equivalent form

12

K m≡ ⋅v v (9.2.2)

Let’s consider a case in which our car changes velocity. For our initial state, the car moves with an initial velocity 0 ,0

ˆxv=v i along the x -axis. For the final state (at some

later time), the car has changed its velocity and now moves with a final velocity

,ˆ

f x fv=v i . Therefore the change in the kinetic energy is

20

1 12 2fK mv mvΔ = − 2 . (9.2.3)

9.2.1 Example: Change in Kinetic Energy of a Car (a) Suppose car A increases its speed from 10 to 20 mph and car B increases its speed from 50 to 60 mph. Both cars have the same mass . What is the ratio of the change of kinetic energy of car

mB to the change of kinetic energy of car A? Which car has a greater

change in kinetic energy? Answer: The ratio of the change in kinetic energy of car B to car A is

2 2 2 2, ,0 , ,2 2

2 2 , ,, ,0

2 2

2 2

1 1( ) ( ) ( ) ( )2 21 1 ( ) ( )( ) ( )2 2(60 mph) (50 mph) 11/ 3.(20 mph) (10 mph)

B f B B f BB

AA f A

m v m v v vKK vm v m v

− −Δ= =

Δ −−

−= =

−

0

0A f Av (9.2.4)

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Thus car B has a much greater increase in its kinetic energy than car A. (b) What is the ratio of the change in kinetic energy of car B to car A as seen by an observer moving with the initial velocity of car A? Answer: Car A now increases its speed from rest to 10 mph and car B increases its speed from 40 to 50 mph. The ratio is now

2 2 2 2, ,0 , ,2 2

2 2 , ,, ,0

2 2

2

1 1( ) ( ) ( ) ( )2 21 1 ( ) ( )( ) ( )2 2(50 mph) (40 mph) 9.

(10 mph)

B f B B f BB

AA f A

m v m v v vKK vm v m v

− −Δ= =

Δ −−

−= =

0

0A f Av (9.2.5)

The ratio is greater than that found in part a). Note that from the new reference frame both car A and car B have smaller increases in kinetic energy. 9.3 Kinematics and Kinetic Energy in One Dimension Constant Acceleration Motion Let’s consider the uniformly accelerated motion of a rigid body in one dimension. We begin the discussion by treating our object as a point mass and ignoring any rotation of the body. Suppose at the body has an initial velocity component in the 0t = x -direction given by ,0xv . If the acceleration is in the direction of the displacement of the body then the body will increase its speed. If the acceleration is opposite the direction of the displacement then the acceleration will decrease the body’s speed. The displacement of the body is given by

20

12 xx v t a tΔ = + . (9.3.1)

The product of acceleration and the displacement is

20

12x x xa x a v t a t⎛Δ = +⎜

⎝ ⎠⎞⎟ . (9.3.2)

The acceleration is given by

( ), ,0x f xx

x

v vvat t

−Δ= =

Δ (9.3.3)

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and therefore

( ) ( ), ,0 , ,0 2

012

x f x x f xx

v v v va x v t t

t t

⎛ ⎞− −⎜Δ = +⎜⎝ ⎠

⎟⎟

. (9.3.4)

Equation (9.3.4) becomes

( )( ) ( ) 2, ,0 0 , ,0 , ,0 , ,

1 1( )2 2x x f x x f x x f x x f xa x v v v v v v v v vΔ = − + − − = − 2

012

. (9.3.5)

If we multiply each side of Equation (9.3.5) by the mass of the body this kinematical result takes on an interesting interpretation for the motion of the object. We have

m

2 2, ,0

1 12 2x x f x f ima x mv m v K KΔ = − = − . (9.3.6)

Recall that for one-dimensional motion, Newton’s Second Law is x xF ma= ; for the situation considered here, Equation (9.3.6) becomes x fF x K KiΔ = − . (9.3.7) Non-constant Acceleration: If the acceleration is not constant, then we can divide the displacement into intervals indexed by

N1 toj N= . It will be convenient to denote the displacement intervals by jxΔ ,

the corresponding time intervals by jtΔ and the velocities at the beginning and end of each interval as , 1x jv − and ,x jv . Note that the velocity at the beginning and end of the first interval, 1j = is then and the velocity at the end of the last interval, 0v j N= is

, ,x N xv v= f .

Consider the sum of the products of the average acceleration ( ), avex ja and displacement

jxΔ in each interval,

( ), ave1

j N

x jj

a=

=jxΔ∑ . (9.3.8)

The average acceleration over each interval is equal to

( ) ( ), , 1, ave

x j x jx j

j

v va

t−−

=Δ

, (9.3.9)

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and so the contribution in each integral can be calculated as above and we have that

( ) 2, ,ave

1 12 2x j j x j x ja x v v 2

, 1−Δ = − . (9.3.10)

When we sum over all the terms only the last and first terms survive, all the other terms cancel in pairs, and we have that

( ) 2, ,ave

1

1 12 2

j N

x j j x N xj

a x v v=

=

Δ = −∑ 2,0 . (9.3.11)

In the limit as and 0 for all N → ∞ jxΔ → j (both conditions must be met!), the limit of the sum is the definition of the integral of the acceleration with respect to the position,

(9.3.12) ( )final

, ave1 initial0

lim .j

j N

x j j xN jx

a x a=

→∞=Δ →

Δ ≡∑ ∫ dx

After multiplying by the mass and using Equation m (9.3.11), which is independent of the limiting process, the integral is equal to the change in kinetic energy,

final

initialx fm a dx K Ki= −∫ . (9.3.13)

It is important to note that Equation (9.3.13) is a mathematical result about displacement, velocity and acceleration, and as of yet has no physical content. When we introduce Newton’s Second Law in the form total

x xF m= a , then

final

total

initial

.x fF dx K K= −∫ i (9.3.14)

The integral of the total x -component of the force with respect to displacement in Equation (9.3.14) applies to the center of mass motion because the total external force on a rigid body causes the center of mass to accelerate (see Section 8.2). 9.4 Work done by Constant Forces We will begin our discussion of the concept of work by analyzing the motion of a rigid body in one dimension acted on by constant forces. Let’s consider an example of this type of motion: pushing a cup forward with a constant force along a desktop. When the cup changes speed (and hence kinetic energy), the sum of the forces acting on the cup

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must be non-zero according to Newton’s Second Law. There are three forces involved in this motion: the applied pushing force appliedF ; the contact force k≡ +C N f ; and gravity,

grav m=F g . The force diagram on the cup is shown in Figure 9.2.

Figure 9.2 Force diagram for cup. Let’s choose our coordinate system so that the x+ -direction is the direction of the forward motion of the cup. The pushing force can then be described by applied applied,

ˆxF=F i . (9.4.1)

Definition: Work done by a Constant Force

Suppose a body moves from an initial point 0x to a final point fx so that the displacement of the cup is positive, 0 0fx x xΔ ≡ − > . The work done by a

constant force applied applied,ˆ

xF=F i acting on the body is the product of the component of the force applied, xF and the displacement xΔ ,

applied applied, xW F x= Δ . (9.4.2)

Work is a scalar quantity; it is not a vector quantity. The SI unit for work is

[ ]2 2 2[1 N m] 1kg m s 1m 1kg m s [1 J]−⎡ ⎤ ⎡ ⎤⋅ = ⋅ ⋅ = ⋅ ⋅ =⎣ ⎦ ⎣ ⎦− . (9.4.3).

Note that work has the same dimension and the same SI unit as kinetic energy. Since our applied force is along the direction of motion, both and . The work done is just the product of the magnitude of the applied force and the distance moved and is positive.

applied, 0xF > 0xΔ >

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We can extend the concept of work to forces that oppose the motion, like friction. In our example of the moving cup, the friction force is k friction k k

ˆ ˆx

ˆf N mμ μ= = = − = −f f i i g i (9.4.4) where from consideration of the N mg= j -components of force in Figure 9.2 and the model k kf Nμ= for kinetic friction have been used. Here the component of force is in the opposite direction as the displacement. The work done by the friction force is negative, friction kW mg xμ= − Δ . (9.4.5) Since the gravitation force is perpendicular to the motion of the cup, the gravitational force has no component along the line of motion. Therefore the gravitation force does zero work on the cup when the cup is slid forward in the horizontal direction. The normal force is also perpendicular to the motion, and hence does no work. We see that the pushing force does positive work, the friction force does negative work, and the gravitation and normal forces do zero work. 9.4.1 Example: Cup on a horizontal table Push a cup of mass 0.20 kg along a horizontal table with a force of magnitude for a distance of . The coefficient of kinetic friction between the table and the cup is

2.0 N0.5 m

k 0.1μ = . Calculate the work done by the pushing force and the work done by the friction force. Answer: The work done by the pushing force is . (9.4.6) applied applied, (2.0 N)(0.50 m) 1.0 JxW F x= Δ = = The work done by the friction force is ( )2

friction k (0.1)(0.2 kg) 9.8 m s (0.50 m)= 0.1 JW mg xμ −= − Δ = − ⋅ − . (9.4.7) Note that the result in (9.4.7) is known to only one significant figure since kμ is given to only one significant figure. However, the precision in (9.4.6) and (9.4.7), , is the same.

0.1J

9.4.2 Example: Cup on a table, applied force at an angle

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Suppose we push the cup in the previous example with a force of the same magnitude but at an angle upwards with respect to the table. Calculate the work done by the pushing force. Calculate the work done by the friction force. The force diagram on the cup is shown in Figure 9.3.

30θ =

Figure 9.3 Force diagram on cup. Solution: The x -component of the pushing force is now . (9.4.8) applied, applied cos (2.0 N)cos(30 ) 1.7 NxF F θ= = = The work done by the pushing force is . (9.4.9) 1

applied applied, (1.7 N)(0.5 m) 8.7 10 JxW F x −= Δ = = × The friction force is k k, k

ˆx

ˆf Nμ= = −f i i . (9.4.10) In this case, the magnitude of the normal force is not simply the same as the weight of the cup. We need to find the y -component of the applied force, ( )applied, applied sin (2.0 N)sin 30 1.0 NyF F θ= = = . (9.4.11) To find the normal force, we apply Newton’s Second Law in the y -direction,

(9.4.12)

( )

applied,

applied,

2 1

0

(0.20 kg) 9.8 m s (1.0 N) 9.6 10 N.

y

y

F N mg

N mg F− −

+ − =

= −

= ⋅ − = ×

The work done by the friction force is then

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( )1friction k (0.1) 9.6 10 N (0.5 m) 4.8 10 JW N xμ −= − Δ = − × = × 2− (9.4.13)

or to one significant figure. Strictly speaking, the result in 25.0 10 J−× (9.4.12) should be rounded to , which would give the same result as 0.1N (9.4.13) to one figure. 9.4.3 Example: Work done by Gravity Near the Surface of the Earth Consider a point-like body of mass near the surface of the earth falling directly towards the center of the earth. The gravitation force between the body and the earth is nearly constant,

m

gravity m=F g . Let’s choose a coordinate system with the origin at the surface of the earth and the + y -direction pointing away from the center of the earth Suppose the body starts from an initial point 0y and falls to a final point fy closer to the earth. How much work does the gravitation force do on the body as it falls? Answer: The displacement of the body is negative, 0 0fy y yΔ ≡ − < . The gravitation force is given by gravity gravity,

ˆym F mg= = = −F ˆg j j . (9.4.14)

The work done on the body is then gravity gravity, yW F y mg y= Δ = − Δ . (9.4.15) For a falling body, the displacement of the body is negative, 0 0fy y yΔ ≡ − < ; therefore the work done by gravity is positive, gravity gravity, 0yW F y mg y= Δ = − Δ > . (9.4.16) The gravitation force is pointing in the same direction as the displacement of the falling object so the work should be positive. When an object is rising while under the influence of a gravitation force,

. The work done by the gravitation force for a rising body is negative, 0 0fy y yΔ ≡ − > gravity gravity, 0yW F y mg y= Δ = − Δ < (9.4.17) because the gravitation force is pointing in the opposite direction from that in which the object is displaced. It’s important to note that the choice of the positive direction as being away from the center of the earth (“up”) does not make a difference. If the downward direction were chosen positive, the falling body would have a positive displacement and the

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gravitational force as given in Equation (9.4.14) would have a positive downward component; the product would still be positive. gravity, yF Δy 9.5 Work and the Dot Product Work is an important physical example of the mathematical operation of taking the dot product between two vectors. Recall that when a constant force acts on a body that is moving along the x -axis, only the component of the force along that direction contributes to the work, xW F x= Δ . (9.5.1) Suppose we are pulling a body along a horizontal surface with a force . Choose coordinates such that horizontal direction is the

Fx -axis and the force forms an angle F

β with the positive x-direction. In Figure 9.4 we show the force vector x yˆ ˆF F= +F i j

and the displacement vector ˆxΔ = Δx i . Note that xΔ is the component of the displacement and hence can be greater, equal, or less than zero (but is shown as greater than zero in the figure for clarity).

Figure 9.4 Force and displacement vectors The dot product between the force vector F and the displacement vector is Δx ( ) ( )ˆ ˆ ˆ

x y xF F x F⋅Δ = + ⋅ Δ = ΔF x i j i x . (9.5.2)

The work done by the force is then W = ⋅ ΔF x . (9.5.3) In general, the angle β takes values within the range π β π− ≤ ≤ (in Figure 9.4,

0 / 2β π− < < ). Since the x -component of the force is cosxF F β= where | |F = F is

the magnitude of F , the work done by the force is

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( )cos cosW F x F xβ β= ⋅Δ = Δ = ΔF x . (9.5.4) 9.6 Work done by Non-Constant Forces Consider a body moving in the x -direction under the influence of a non-constant force in the x -direction, x

ˆF=F i . The body moves from an initial position 0x to a final position

fx . In order to calculate the work done by a non-constant force, we will divide up the displacement into a large number of small displacements N jxΔ where the index j

marks the thj displacement and takes integer values from 1 to , as in Section 9.3. Let denote the average value of the

N

( ), avex jF x -component of the force in the displacement

interval 1j jx ,x−⎡⎣ ⎤⎦ . For the thj displacement interval we calculate the contribution to the work ( ), avej x jW F xjΔ = Δ

jx

. (9.6.1)

This contribution is a scalar so we add up these scalar quantities to get the total work

( ), ave1 1

j N i N

N j x jj i

W W F= =

= =

= Δ = Δ∑ ∑ . (9.6.2)

The sum in Equation (9.6.2) depends on the number of divisions and the width of the intervals

NjxΔ . In order to define a quantity that is independent of the divisions, we take

the limit as and N → ∞ 0jxΔ → . The work is then

. (9.6.3) ( )0

, ave10

limf

j

x xj N

x j j xN j x xx

W F x==

→∞= =Δ →

= Δ =∑ ∫ F dx

This last expression is the definition of the integral of the x-component of the force with respect to the parameter x . In Figure 9.5 we graph the x -component of the force as a function of the parameter x . The work integral is the area under this curve between

0x x= and fx x= .

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Figure 9.5 Graph of x -component of a sample force as a function of the parameter x . 9.6.1 Example: Work done by the Spring Force Connect one end of a spring to a body resting on a smooth (frictionless) table and fix the other end of the spring to a wall. Stretch the spring and release the spring-body system. How much work does the spring do on the body as a function of the stretched or compressed length of the spring? Answer: We first begin by choosing a coordinate system with origin at the position of the body when the spring is at rest in the equilibrium position. We choose the ˆ unit vector to point in the direction the body moves when the spring is being stretched and the coordinate

i

x to denote the position of the body with respect to the equilibrium position, as in Figure 9.6 (which indicates that in general the position x will be a function of time). The spring force on the body is given by x

ˆF k= = −F i ˆx i . (9.6.4)

Figure 9.6 Equilibrium position and position at time t In Figure 9.7 we show the graph of the x -component of the spring force as a function of x for both positive values of x corresponding to stretching, and negative values of x corresponding to compressing of the spring. Note that 0x and fx can be positive, zero, or negative.

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Figure 9.7 The x -component of the spring force as a function of x . The work done is just the area under the curve for the interval 0x to fx ,

. (9.6.5) 0 0

( )f fx x x x

xx x x x

W F dx k x= =

= =

= = −∫ ∫ dx

This integral is straightforward; the work done by the spring force on the body is

(0

2 20

1( )2

fx x

fx x

W k x dx k x=

=

= − = − −∫ )x . (9.6.6)

When the absolute value of the final distance is less than the absolute value of the initial distance, 0fx x< , the work done by the spring force is positive. This means that if the spring is less stretched or compressed in the final state than in the initial state, the work done by the spring force is positive. The spring force does positive work on the body when the spring goes from a state of “greater tension” to a state of “lesser tension.” We shall see in Chapter 10 that the positive work done by the spring force decreases the potential energy stored in the spring. 9.6.2 Example: Work done by the Inverse Square Gravitation Force Consider a body of mass moving directly towards the sun, which has mass . Let’s choose a coordinate system with the origin at the center of the sun. Choose the -axis to be directed along the motion of the body with the r unit vector pointing away from the center of the sun. Initially the body is at a distance from the center of the sun. In the final configuration the body has moved to a distance

m sunmr

0r

f 0r r< from the center of the sun. How much work does the gravitation interaction between the sun and the body do on the body during this motion?

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Answer: The gravitation force between the sun and the body is given by

sg g 2

Gm mˆFr

= = −F r r . (9.6.7)

Figure 9.8 Graph of the radial force between the sun and the body as a function of the separation . The vertical and horizontal scales are in arbitrary units. r

The work done by this gravitation force on the body is given by the integral

0 0

sungrav 2

f fr r

r r

Gm mW F dr drr

⎛= = −⎜⎝ ⎠∫ ∫ ⎞

⎟ . (9.6.8)

Upon evaluation of this integral, we have for the work

00

sun sunsun2

0

1 1f fr r

r fr

Gm m Gm mW dr Gm mr r r

⎛ ⎞⎛ ⎞= − = = −⎜⎜ ⎟ ⎜⎝ ⎠ ⎝ ⎠∫ r

⎟⎟ . (9.6.9)

Since the body has moved closer to the sun, 0fr r< , hence . Thus the work done by gravitation force between the sun and the body on the body is positive,

01/ 1/fr > r

sun0

1 1 0f

W G m mr r

⎛ ⎞= − >⎜ ⎟⎜ ⎟

⎝ ⎠. (9.6.10)

We expect this result because the gravitation force points along the direction of motion, so the work done by the force is positive. Also we expect that the sign of the work is the same for a body moving closer to the sun as a body falling towards the earth in a constant gravitation field, as seen in Example 9.4.3 above. We shall see in Chapter 10 that the positive work done by the gravitation force will decrease the potential energy stored in the gravitation field.

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9.6.3 Example: Work Done by the Inverse Square Electrical Force Let’s consider two point-like bodies, body 1 and body 2, with charges and respectively interacting via the electric force alone. Body 1 is fixed in place while body 2 is free to move. Let’s choose a coordinate system with the origin at body 1. Let denote the distance between body 2 and the origin. Choose the

1q 2q

rˆ+r unit vector to be directed

radially away from the origin. Initially a distance separates the bodies and in the final state a distance

0r

fr separates the bodies. If the charges of the bodies are of the same sign they will repel and . If the charges of the bodies are of opposite signs, the bodies will attract and . How much work does the electric force do on body 2 during this motion?

0fr r>

0fr r<

Answer: This problem is nearly identical to computing the gravitation work done on a body as it moves towards another body. The most significant difference is that the electric force can be either attractive or repulsive while the gravitation force is always attractive. The electric force between the bodies is given by

1 2elec elec 2

0

14r

q qˆ ˆF Frπε

= = =F r r r . (9.6.11)

Figure 9.9 Graph of the electric force between the two charges of the same sign as a function of the separation r . The vertical and horizontal scales are in arbitrary units.

The work done by this electric force on the body 2 is given by the integral

0 0

1 22

0

14

f fr r

rr r

q qW F dr drrπε

= =∫ ∫ . (9.6.12)

Evaluating this integral, we have for the work done by the electric force

6/26/2008 - 17 -

0 0

1 2 1 21 22 2

0 0 0

1 1 14 4 4

ff rr

fr r

q q q qW dr q qr r rπε πε πε

⎛ ⎞= = − = − ⎜⎜

⎝ ⎠∫

0

1 1r

− ⎟⎟ . (9.6.13)

If the bodies have the opposite signs, 1 2 0q q < , we expect that the body 2 will move closer to body 1 so , and . From our result for the work, the work done by electrical force in moving body 2 is positive,

0fr r< 01/ 1/fr > r

1 20 0

1 1 1 04 f

W q qr rπε

⎛ ⎞= − −⎜ ⎟⎜ ⎟

⎝ ⎠>

r

. (9.6.14)

Once again we see that bodies under the influence of electric forces only will naturally move in the directions in which the force does positive work. As the body 2 moves closer to body 1, the stored potential energy between the charges decreases, as will be shown in Chapter 10. If the bodies have the same sign, . The bodies will repel, with the result

and . Thus the work is once again positive: 1 2 0q q >

0fr r> 01/ 1/fr <

1 20 0

1 1 1 04 f

W q qr rπε

⎛ ⎞= − −⎜ ⎟⎜ ⎟

⎝ ⎠> . (9.6.15)

9.7 Work done along an Arbitrary Path Now suppose that a non-constant force F acts on a point-like body of mass while the body is moving on a three-dimensional curved path. The position vector of the body at time with respect to a choice of origin is

m

t ( )tr .2 In Figure 9.10 we show the orbit of the body for a time interval 0[ , ]ft t moving from an initial position 0 (t t0 )≡ =r r at time 0t t= to a final position ( )f ft t≡ =r r at time ft t= .

We divide the time interval 0[ , ]ft t into smaller intervals , N 1j jt ,t−⎡ ⎤⎣ ⎦ 1j N=

with . Consider two position vectors Nt t= f ( )j jt t≡ =r r and ( )1j t t 1j− −≡ =r r and denote

the displacement vector during the corresponding time interval as 1j j j−Δ = −r r r .

2 We use the parameter t and refer to this parameter as “time” to avoid confusion. The fact that the work done is independent of parameterization is a standard result of vector calculus, and need not be rederived in these notes. Examples are given in the Appendix to this chapter.

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Figure 9.10 Path traced by the motion of a body. Let denote the average force acting on the body during the interval .

The average force in this interval is iF 1[ ,j jt t− ]

( )avejF and he average work jWΔ done by the force

during the time interval is the dot product between the average force vector and the displacement vector,

1[ ,j jt t− ]

( )

avej jW jΔ = ⋅F Δr . (9.7.1)

The force and the displacement vectors for the time interval are shown in Figure 9.11 (note that the figure uses “ i ” as index instead of “

1[ ,j jt t− ]j ” and the subscript “ave” on

has been suppressed). ( )avejF

Figure 9.11 An infinitesimal work element. We calculate the work by adding these scalar contributions to the work for each interval , for to , 1[ ,j jt t− ] 1i = N

6/26/2008 - 19 -

( )ave

1 1

j N j N

N i jj j

W W= =

= =j= Δ = ⋅Δ∑ ∑ F r . (9.7.2)

We would like to define work in a manner that is independent of the way we divide the interval, so we take the limit as and N → ∞ 0jΔ →r . In this limit, as the intervals become smaller and smaller, the distinction between the average force and the actual force vanishes. Thus if this limit exists and is well defined, then the work done by the force is

( )0ave

10

lim f

j

j N r

j j rN j

W=

→∞=Δ →

= ⋅Δ =∑ ∫r

F r F d⋅ r . (9.7.3)

Notice that this summation involves adding scalar quantities. This limit is called the line integral of the force . The symbol F d r is called the infinitesimal vector line element. At time , is tangent to the orbit of the body and is the limit of the displacement vector

as approaches zero. In this limit, the parameter t does not appear in the expression in

t d r( ) (t t tΔ = + Δ −r r r ) tΔ

(9.7.3). In general this line integral depends on the particular path the body takes between the initial position and the final position 0r fr , which matters when the force is non-constant in space, and when the contribution to the work can vary over different paths in space. An example is given in the

F

Appendix to this chapter. We can represent the integral in Equation (9.7.3) explicitly in a coordinate system by specifying the infinitesimal vector line element d r and then explicitly computing the dot product. For example, in Cartesian coordinates the line element is ˆ ˆ ˆd dx dy dz= + +r i j k , (9.7.4) where , and represent arbitrary displacements in the dx dy dz x -, y -, and -directions respectively as seen in Figure 9.12.

z

Figure 9.12 A line element in Cartesian coordinates. The force vector can be represented in vector notation by

6/26/2008 - 20 -

x y z

ˆ ˆ ˆF F F= + +F i j k . (9.7.5) The infinitesimal work is the dot product

( ) ( )ˆ ˆ ˆ ˆˆ ˆ

.x y z

x y z

dW d F F F dx dy dz

F dx F dy F dz

= ⋅ = + + ⋅ + +

= + +

F r i j k i j k (9.7.6)

The total work is

(9.7.7) ( )0 0 0 0 0

f f f f f

x y z x y zW d F dx F dy F dz F dx F dy F dz= = = = =

= = = = =

= ⋅ = + + = + +∫ ∫ ∫ ∫ ∫r r r r r r r r r r

r r r r r r r r r r

F r

Equation (9.7.7) shows that W consists of three separate integrals. In order to properly calculate these integrals, we need to know the specific path the body takes. Often we are given an orbit equation for the path, satisfied by the position components of the body. For example, the orbit may be a circle in the x y− plane centered at the origin with radius R , and the position components satisfy the equation 2 2 2x y R+ = . (9.7.8) Another example is the parabolic trajectory associated with projectile motion with the orbit equation in that case is given by

2

,0 ,0 02 002 2 2

,0 ,0 ,0 ,0 ,0

( )2 2

y y

x x x x x

v v xg gx gx0y x x xv v v v v

⎛ ⎞= − + + − − +⎜ ⎟⎜ ⎟

⎝ ⎠y . (9.7.9)

A common difficulty in calculating line integrals is to explicitly determine how the force varies over the trajectory. In a more general case we must consider the possibility that each component of the force depends on the other coordinates, ( ) ( ) ( )x x y y z zF F x, y,z , F F x, y,z , F F x, y,z= = = (9.7.10) (an example is given in the Appendix to this chapter). If the force depends on the velocity of the body, the position of the body must be known as a function of time as well as space. 9.7.1 Example: Work Done in a Constant Gravitation Field

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The work done in a uniform gravitation field is a fairly straightforward calculation when the body moves in the direction of the field. Suppose the body is moving under the influence of gravity, ˆmg= −F j along a parabolic curve (see Section 5.8) given by the equation ( ) 2y y x a x b x c= = − + . (9.7.11) The body begins at the point 0 0( , )x y and ends at the point ( , )f fx y . What is the work done by the gravitation force on the body? Answer: The position vector now specifies the path ( ) 2ˆ ˆ ˆ( ) ( ) ˆx x y x x ax bx c= = + = + − +r r i j i j . (9.7.12) The infinitesimal line element is therefore dr ( )ˆd dx a 2bx dx= + −r i j . (9.7.13) The dot product that appears in the line integral can now be calculated, . (9.7.14) ˆ ˆ ˆ[ ( 2 ) ] ( 2 )d mg dx a bx dx mg a bx dx⋅ == − ⋅ + − = − −F r j i j The parameter x varies from 0x to fx . The work integral is then

0 0

( 2 )f fr x x

r x x

W d mg a bx=

=

= ⋅ = − −∫ ∫F r dx . (9.7.15)

This integral is fairly straightforward; we integrate and find the work to be

00

2

2 20 0

( 2 ) ( )

( ) ( )

ff

xx

xx

f f

W mg a bx dx mg ax bx

mg b x x a x x .

= − − = − +

⎡ ⎤= − − −⎣ ⎦

∫ (9.7.16)

Because the orbit equation of the body is a parabola,

2

0 0 0

2f f f

y a x b x c

y a x b x c.

= − +

= − + (9.7.17)

6/26/2008 - 22 -

Subtracting the second equation in (9.7.17) from the first (watch the signs) yields ( ) ( ) ( )2 2

0 0f f fb x x a x x y y− − − = − − 0 (9.7.18) The work done by the uniform gravitation force is then

( )0

0f

fW d mg y= ⋅ = − −∫r

rF r y . (9.7.19)

This result is not surprising since the force is only in the y -direction. Therefore the only non-zero contribution to the work integral is in the y -direction, with the result that

. (9.7.20) ( ) (0

0 0

0

f ff

y y y y

yy y y y

W d F dy mg dy mg y= =

= =

= ⋅ = = − = − −∫ ∫ ∫r

rF r )f y

In this case of a constant force, the work integral is independent of path. 9.8 Work-Kinetic Energy Theorem There is a direct connection between the total work done on a body and the change in kinetic energy the body undergoes. If the total work done on the body is non-zero, this implies that an unbalanced force has acted it the body, and the body will have undergone acceleration. We have already shown for one-dimensional motion (Equation (9.3.14))

final

total

initial

.x fF dx K K= −∫ i (9.8.1)

The integral on the left-hand-side of Equation (9.8.1) is the total work done by the applied total force. When the total work done on a body is positive, the body will increase its speed, and negative work done on a body causes a decrease in speed. When the total work done is zero, the body will maintain a constant speed. In fact, the work-energy relationship is quite precise; the total work done by the net applied force on a body is identically equal to the change in kinetic energy of the body,

totalW

2total 0

1 12 2fW K mv m= Δ = − 2v . (9.8.2)

9.8.1 Example: Gravity and the Work-Energy Theorem

6/26/2008 - 23 -

Suppose a ball of mass starts from rest at a height 0.20 kgm = 0 15 0my .= above the surface of the earth and falls down to a height 5.0 mfy = above the surface of the earth. What is the change in the ball’s kinetic energy? Find the final velocity using the work-energy theorem. Answer: As only one force acts on the ball, the change in kinetic energy is the work done by gravity,

(9.8.3) ( )

( )( )( )grav 0

1 22 0 10 kg 9 8m s 5 0m 15 0m 2 0 10 J.

fW mg y y

. . . . .− −

= − −

= − × ⋅ − = × 1

The ball started from rest, , so the change in kinetic energy is ,0 0yv =

2 2, ,0

1 1 12 2 2y f y y fK mv mv mvΔ = − = 2

, . (9.8.4)

We can solve (9.8.4) for the final velocity, using (9.8.3)

( )1

grav 11

2 2 0 10 J22 1 4 10 m s2 0 10 kgy , f

.WKv .m m .

−−

×Δ= = = = ×

×1⋅ . (9.8.5)

For the falling ball in a constant gravitation field, the positive work of the gravitation force on the body corresponds to an increasing kinetic energy and speed. For a rising body in the same field, the kinetic energy and hence the speed decrease since the work done is negative. 9.8.2 Example: Final Kinetic Energy of Moving Cup A person pushes a cup of mass 0.20 along a horizontal table with a force of magnitude 2.0 N at an angle of 30 with respect to the horizontal for a distance of

, as in Example 9.4.1. The coefficient of friction between the table and the cup is

kg

0 50m.

k 0 1.μ = . If the cup was initially at rest, what is the final kinetic energy of the cup after being pushed ? What is the final speed of the cup? 0 50m. Answer: The total work done on the cup is the sum of the work done by the pushing force and the work done by the friction force, as given in Equations (9.4.9) and (9.4.13),

6/26/2008 - 24 -

( )( )

( )total applied friction applied, 0

21.7 N 9.6 10 N (0.5 m) 8.0 10 J.

x k fW W W F N x xμ−

= + = − −

= − × = × 1− (9.8.6)

According to our work-kinetic energy theorem,

2total 0

1 12 2fW K mv m= Δ = − 2v . (9.8.7)

The initial velocity is zero so the change in kinetic energy is just

2 2, ,0

1 1 12 2 2y f y y fK mv mv mvΔ = − = 2

, . (9.8.8)

Thus the work-kinetic energy theorem enables us to solve for the final kinetic energy,

2total

1 8.0 10 J2f fK mv K W −= = Δ = = × 1 (9.8.9)

We can solve for the final speed,

( )1

1total,

2 8.0 10 J2 2 2.9 m s0.2 kg

fy f

K Wvm m

−−

×= = = = ⋅ . (9.8.10)

9.9 Power Applied by a Constant Force Suppose that an applied force appliedF acts on a body during a time interval and displaces the body in the

tΔx -direction by an amount xΔ . The work WΔ done during this

interval is applied, xW F xΔ = Δ . (9.9.1) where applied, xF is the x -component of the applied force. (Equation (9.9.1) is the same as Equation (9.4.2).) The average power of this applied force is defined to be the rate at which work is done, so that

applied,ave applied, ,ave

xx x

F xWPt t

F vΔΔ

= = =Δ Δ

. (9.9.2)

6/26/2008 - 25 -

The average power delivered to the body is equal to the product of the component of the force in the direction of motion and the component of the average velocity of the body. Power is a scalar quantity and can be positive, zero, or negative depending on the sign of work. The SI unit of power the watt [ ; W] 1[1 W] 1 J s−⎡ ⎤= ⋅⎣ ⎦ . Definition: Instantaneous Power

The instantaneous power at time is defined to be the limit of the average power as the time interval [ approaches zero,

t, ]t t t+ Δ

applied,applied, applied,0 0 0

lim lim limxx x xt t t

F xWP Ft t tΔ → Δ → Δ →

x F vΔΔ ⎛ ⎞= = = =⎜ ⎟Δ Δ Δ⎝ ⎠

Δ . (9.9.3)

The instantaneous power of a constant applied force is the product of the component of the force in the direction of motion and the instantaneous velocity of the moving object. A more general result is found by using the expression for as given in Equation dW(9.7.6), dW d= ⋅F r (9.9.4)

dW dPdt dt

⋅= = =

F r F v⋅ . (9.9.5)

9.9.1 Example: Gravitational Power for a Falling Mass Suppose a ball of mass starts from rest at a height 0.20 kgm = 0 15.0 my = above the surface of the earth and falls down to a height 5.0 mfy = above the surface of the earth. What is the average power exerted by the gravitation force? What is the instantaneous power when the ball is at a height 5.0 mfy = above the surface of the earth? Make a graph of power as a function of time. Ignore the effects of air resistance. Answer: There are two ways to solve this problem. Both approaches require calculating the time

ft for the ball to fall. Set for the time the ball was released. We can solve for the time interval

0 0t =

ft tΔ = that it takes the ball to fall using the equation for a freely falling object that starts from rest,

20

12f fy y gt= − . (9.9.6)

6/26/2008 - 26 -

Thus the time interval for falling is

( ) ( )0 2

2 2 15.0 m 5.0 m 1.4 s9.8 m sf ft y y

g −= − = − =⋅

. (9.9.7)

First Approach: We can calculate the work done by gravity,

(9.9.8) ( )

( )( )grav 0

1 22.0 10 kg 9.8 m s (5.0 m 15.0 m) 2.0 10 J.

fW mg y y− −

= − −

= − × ⋅ − = × 1

Then the average power is

1

1ave

2.0 10 J 1.4 10 W1.4 s

WPt

Δ ×= = = ×

Δ. (9.9.9)

Second Approach: We calculate the gravitation force and the average velocity. The gravitation force is ( )( )1 2

grav, 2.0 10 kg 9.8 m s 2.0 NyF mg − −= − = − × ⋅ = − . (9.9.10) The average velocity is

1,ave

5.0 m 15.0 m 7.0 m s1.4 sy

yvt

−Δ −= = = − ⋅

Δ. (9.9.11)

The average power is therefore

(9.9.12) ( )ave ,ave ,ave

1

( )

( 2.0 N) 7.0 m s 1.4 10 W.y y yP F v mg v

−

= = −

= − − ⋅ = × 1

In order to find the instantaneous power at any time, we need to find the instantaneous velocity at that time. The ball takes a time 1.4 sft = to reach the height . The velocity at that height is given by

5.0 mfy =

( )29.8 m s (1.4 s) 1.4 10 m sy fv g t 1 1− −= − = − ⋅ = − × ⋅ . (9.9.13)

6/26/2008 - 27 -

The instantaneous power at time 1.4 sft = is then

( )

( )

2

22 1

( )

(0.2 kg) 9.8 m s (1.4 s) 2.7 10 W.

y y f fP F v mg gt mg t

−

= = − − =

= ⋅ = × (9.9.14)

If this problem were done symbolically, the answers given in (9.9.12) and (9.9.14) would differ by a factor of two; the answers have been rounded to two significant figures. The instantaneous power grows linearly with time so the graph of power vs. time is shown in Figure 9.13. From the figure, it should be seen that the instantaneous power at any time is half of the average power between 0t = and that time.

Figure 9.13 Graph of power vs. time. 9.9.2 Example: Power Pushing a Cup A person pushes a cup of mass 0.20 along a horizontal table with a force of magnitude 2.0 N at an angle of 30 with respect to the horizontal for a distance of

, as in Example 9.4.1. The coefficient of friction between the table and the cup is

kg

0 50m.

k 0 1.μ = . What is the average power of the pushing force? What is the average power of the kinetic friction force? Answer: We will use the results from Example 9.8.2 above, but keeping extra significant figures in the intermediate calculations. The work done by the pushing force is . (9.9.15) 1

applied applied, 0( ) (1.732 N)(0.50 m) 8.660 10 Jx fW F x x −= − = = × The final speed of the cup is 12 860m sx , fv . −= ⋅ . Assuming constant acceleration, the time during which the cup was pushed is

6/26/2008 - 28 -

( )02

0 3496sff

x , f

x xt

v−

= = . . (9.9.16)

The average power of the pushing force is then, with ft tΔ = ,

( )1

appliedapplied ave

8.660 10 J 2.340 W0.3496 s

WP

t

−Δ ×= = =

Δ, (9.9.17)

or 2 to two significant figures. 3W. The work done by the friction force is

(9.9.18) ( )

( ) (friction k 0

2 2k 0( ) 9.6 10 N (0.50 m) 4.8 10 J .

f

f

W f x x

N x xμ − −

= −

= − − = − × = − × ) The average power of kinetic friction is

( )2

1frictionfriction ave

4.8 10 J 1.373 10 W0.3496 s

WPt

−−Δ − ×

= = = − ×Δ

(9.9.19)

or to two significant figures. 11.4 10 W−− × Time Rate of Change of Kinetic Energy and Power The time rate of change of the kinetic energy for a body of mass moving in the m x -direction is

212

xx x x

dK d dvmv m v ma vdt dt dt

⎛ ⎞= = =⎜ ⎟⎝ ⎠

x . (9.9.20)

By Newton’s Second Law, x xF ma= , and so Equation (9.9.20) becomes

x xdK F v Pdt

= = ; (9.9.21)

the instantaneous power delivered to the body is equal to the time rate of change of the kinetic energy of the body. The result found above is more general; applied to bodies moving in arbitrary directions, differentiation of Equation (9.2.2) gives

6/26/2008 - 29 -

12

dK d m mdt dt

⎛ ⎞= ⋅ = ⋅ = ⋅⎜ ⎟⎝ ⎠

v v a v F v P= , (9.9.22)

consistent with Equation (9.9.5).

6/26/2008 - 30 -