6-1

Chapter 6 Conservation of Energy

Today we begin with a very useful concept – Energy.

We will encounter many familiar terms that now have very specific definitions in physics.

Conservation of energy

Work

Potential

Power

In some cases, it can be argued that these terms have a physics definition that is similar to its

everyday usage.

The Law of Conservation of Energy

The total energy in the universe is unchanged by any physical process:

total energy before = total energy after

From page 187: “In ordinary language, conserving energy means trying not to waste useful

energy resources. In the scientific meaning of conservation, energy is always conserved no

matter what happens.”

Conservation of energy is one of the few universal principles of physics. No exception has ever

been found. It applies to physical, chemical, and biological systems.

Also from page 187: “Some problems can be solved using either energy conservation or

Newton’s second law. Usually the energy method is easier.” Using Newton’s second law

involves vector methods since forces are vector quantities. Much of the time, energy involves

scalar quantities, which are much easier to deal with (and more familiar). “When deciding which

of these two approaches to use to solve a problem, try using energy conservation first.”

Kangaroos are mentioned at the beginning of this chapter.

http://www.youtube.com/watch?v=hijYSR2MFiY

Forms of Energy

At the most fundamental level there are three kinds of energy

1. Kinetic energy – energy due to motion

2. Potential energy – energy due to

interaction

a. Gravitational potential energy –

interaction between the Earth

and a mass

b. Elastic potential energy –

interaction between a spring and

a mass

3. Rest energy – internal energy to a body

Energy is measured in Joules (J).

6-2

Work

Suppose a force F

causes an object to move a distance x parallel to F

The work done by a constant force F

is defined as

xFW

DO NOT MEMORIZE!

Suppose a constant force F

causes an object to move along r

not parallel to F

x

6-3

cosrFW

where is the angle between F

and r

. MEMORIZE.

Work is a scalar quantity and can be positive, negative, or zero.

Positive: between 0o and 90o

Negative: between 90o and 180o

Zero: = 90o

o Usually tension and normal force do no work. The exception is (c) below.

The work done by several forces can be found from the net force

cos

21

rF

WWWW

net

Ntotal

Work and Kinetic Energy

Choosing the x axis along Fnet, (using x = r cos )

xma

xFW

x

nettotal

We had an equation from Chapter 2

6-4

)(

2

22

21

22

ixfxx

xixfx

vvxa

xavv

Substituting into Wtotal

)(22

21

ixfxtotal vvmW

Since the net force is in the x-direction, ay and az are both zero. Only the x-component of the

velocity changes

2222222222

)()( ixfxiziyixfzfyfxif vvvvvvvvvv

and

)(22

21

iftotal vvmW

The translational kinetic energy is defined as

2

21 mvK

The work-kinetic energy theorem is

KWtotal

While this expression is foundational to this chapter, do not memorize. We shall derive a more

useful form.

Gravitational Potential Energy (1)

The weight can do work. Toss a ball up and it slows down. In our new language, its kinetic

energy decreases. The kinetic energy is converted into another form of energy we call

gravitational potential energy.

The change in gravitational potential energy

gravgrav WU

In terms of position

ymgUgrav

This equation is true even if the object does not travel in a straight line.

6-5

The gravitational potential energy is

mgyUgrav

The final form of our relation is

UKWnc

This is it. You need to know it. We have another entry into our cause and effect table.

Another useful form is

ffncii

ifif

nc

UKWUK

UUKK

UKW

Wnc is the work done by nonconservative forces. Nonconservative forces do not have a potential

energy. A good example is friction.

The mechanical energy is

UKE

Conservation of Mechanical Energy

When nonconservative forces do no work, mechanical energy is conserved:

fi EE

6-6

The zero of potential energy is arbitrary. Choose whatever is convenient, usually the ground.

The work done by a conservative force is independent of the path taken.

Problem: A 0.1-kg ball is thrown at 5 m/s from a 10 m tower. What is its speed when it is 5 m

above the ground? What is its speed when it hits the ground?

Solution: Use the conservation of mechanical energy.

m/s1.11

)m5m10(m/s8.9(2)m/s5(

)(2

22

21

2

12

2

221

2

2

121

1

2

221

2

2

121

1

2211

21

yygvv

vgyvgy

mvmgymvmgy

KUKU

EE

At the bottom, y2 = 0, v2 = 14.9 m/s.

Notice the direction of the throw is not mentioned. No matter which way the ball is thrown, it

has the same speed at the same height! This is very hard to prove using Newton’s second law.

Gravitational Potential Energy (2)

For objects far from the Earth,

r

GmMU E

While this looks very different from mgy, the text (p. 203) shows they are equivalent.

Example 6.8 What is the escape velocity for the Earth?

Solution: Use conservation of energy. We cannot use U = mgy for an object far from the earth!

2

212

21

f

f

E

i

i

E

ffii

fi

mvr

Gmmmv

r

Gmm

KUKU

EE

When an object escapes from the Earth, Earth’s gravity is not acting on it and rf → ∞. If the

object barely escapes the Earth, vf = 0.

6-7

i

E

i

i

i

E

r

Gmv

mvr

Gmm

2

02

21

The starting position is on the Earth’s surface and ri = RE = 6.36×106 m. The mass of the Earth is

mE = 5.97×1024 kg. The escape velocity is

m/s200,11)kg1037.6(

)kg1097.5)(kg/Nm1067.6(226

242211

i

E

ir

Gmv

This is

s

mi0.7

m1609

mi1

s

m200,11

iv

To summarize, we developed what is usually called the work-energy theorem:

UKWnc or ffncii UKWUK

where K is the kinetic energy

2

21 mvK ,

U is the potential energy (usually gravitational)

mgyUgrav

and Wnc is the total work performed by all nonconservative forces. In many situations (but not

all), mechanical energy is conserved and Wnc = 0.

Work done by a variable force.

The advantage of energy methods is seen when dealing with a variable force. Suppose the force

changes with displacement. An example of such a device is shown on the next page. How do

we calculate the work done?

We divide the overall displacement into a series of small displacements, x. Over each

of the smaller displacements, the force is almost constant. The work done for a small

displacement is

xFW x

6-8

The total work is the area under the curve shown above. This procedure was used to find the

gravitational potential energy for an object far from the Earth

r

mGMU E

Hooke’s Law and Ideal Springs

The force exerted by the archer

increases as the bowstring is drawn

back. Robert Hooke proposed an ideal

spring where the force is proportional

to the displacement

kxFx

The displacement of the spring from the

relaxed position is x. The constant k, is

called the spring constant. It is

measured in N/m and it gives the

strength of the spring. The larger k is,

the stiffer the spring. The minus sign

indicates that if the spring is stretched

to the right, the spring pulls back to the

left and vice versa.

6-9

Elastic Potential Energy

As the spring is pulled (or pushed) from its relaxed position, work is done on it. The work done

is independent of the path taken and accordingly, a potential energy can be defined. The elastic

potential energy is found to be

2

21 kxUelastic

Note that U = 0 when x = 0.

Power

Often the rate of energy conversion is important. We use the term power to refer to the rate of

energy conversion. Over an extended time, the average power is

t

EPav

Power is measured in Joules/second or watts (W). Be careful with W for work and W for watts.

Remember that work changes the mechanical energy of the system.

cos

cos

Fv

t

rF

t

EP

Problem 82. A spring gun (k = 28 N/m) is used to shoot a 56-g ball horizontally. Initially the

spring is compressed by 18 cm. The ball loses contact with the spring and leaves the gun when

the spring is still compressed by 12 cm. What is the speed of the ball when it hits the ground 1.4

m below the gun?

Solution: This appears to be a projectile problem. It is an energy problem with two potential

energies. Take the initial position to be at the top and the final position just before it hits the

ground,

6-10

m/s04.6

)kg056.0/())m12.0()m18.0)((N/m28()m4.1)(m/s8.9(2

/)(2

00

222

2

2

2

112

2

2212

221

1

2

121

2211

mxxkgyv

kxmvmgykx

UKUK

Problem 91. A 1500-kg car coasts in neutral down a 2.0º hill. The car attains a terminal speed

of 20.0 m/s. (a) How much power must the engine deliver to drive the car on a level road at 20.0

m/s? (b) If the maximum useful power that can be delivered by the engine is 40.0 kW, what is

the steepest hill the car can climb at 20.0 m/s?

Solution: At the terminal speed, the x-component of the weight of the car is opposed by air

resistance. When coasting down the hill, the free body diagram is

For the x-component (along the incline)

N513

2sin)m/s8.9)(kg1500(

sin

0sin

2

mgF

Fmg

maF

air

air

xx

(a) Now the car is moving along a horizontal road at constant velocity. From the FBD found

at the top of the next page:

airmotor

airmotor

xx

FF

FF

maF

0

6-11

The power delivered by the engine when the car moves at 20 m/s (notice the angle below is not

2º!) is

W300,100cos)m/s20)(N513(0coscos vFvFP airmotormotor

(b) Again a free body diagram is helpful.

Climbing with constant speed, ax = 0,

sin

0sin

mgFF

FmgF

maF

airmotor

airmotor

xx

The maximum power supplied by the engine is 40.0 kW. The power is

sin

0cos)sin(

cos

mgvvF

vmgF

vFP

air

air

motor

6-12

101.0

)m/s20)(m/s8.9)(kg1500(

)m/s20)(N513(W100.40

sin

2

3

mgv

vFP air

This corresponds to a 5.8º angle.