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).
6-1

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
W

F
 x
 x
DO NOT MEMORIZE!

Suppose a constant force F causes an object to move along

 r not parallel to

F
6-2

 
W

F
 r cos
 where

is the angle between

F and

 r . MEMORIZE.
Work is a scalar quantity and can be positive, negative, or zero.

Positive:

between 0 o
and 90 o

Negative:

between 90 o
and 180 o

Zero:

= 90 o 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
W total


W
1

W
2
F net
 r
  
W
N cos

Work and Kinetic Energy
Choosing the x axis along F net
, (using
 x =
 r cos

)
W total


F net
 x ma x
 x
We had an equation from Chapter 2
6-3

v fx
2  v ix
2 a x
 x


2 a x
 x
1
2
( v fx
2  v ix
2
)
Substituting into W total
W total

1
2 m ( v fx
2  v ix
2
)
Since the net force is in the x -direction, a y
and a z
are both zero. Only the x -component of the velocity changes v f
2  v i
2 
( v fx
2  v fy
2  v fz
2
)

( v ix
2  v iy
2  v iz
2
)
 v fx
2  v ix
2 and
W total

1
2 m ( v f
2  v i
2
)
The translational kinetic energy is defined as
K

1
2 mv
2
The work-kinetic energy theorem is
W total
 
K
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

U grav
 
W grav
In terms of position

U grav
 mg
 y
This equation is true even if the object does not travel in a straight line.
6-4

The gravitational potential energy is
U grav
 mgy
The final form of our relation is
W nc
 
K
 
U
This is it. You need to know it.
We have another entry into our cause and effect table.
Another useful form is
K i

U i
W nc

W nc




K
 
U

K f

K i
 
U f

U i

K f

U f
W nc
is the work done by nonconservative forces. Nonconservative forces do not have a potential energy. A good example is friction.
The mechanical energy is
E

K

U
Conservation of Mechanical Energy
When nonconservative forces do no work, mechanical energy is conserved:
E i

E f
6-5

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.
U
1
E
1

K
1

E
2

U
2

K
2 mgy
1

1
2 mv
1
2 gy
1

1
2 v
1
2
 mgy
2

1
2 mv
2
2
 gy
2

1
2 v
2
2 v
2
 v
1
2 
2 g ( y
1
 y
2
)

( 5 m/s )
2

11 .
1 m/s

2 ( 9 .
8 m/s
2
( 10 m

5 m )
At the bottom, y
2
= 0, v
2
= 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,
U
 
GmM
E r
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!
U i

E i
K i

E f

U f

K f

Gmm
E r i

1
2 mv i
2  
Gmm
E r f

1
2 mv f
2
When an object escapes from the Earth, Earth’s gravity is not acting on it and r f
→ ∞. If the object barely escapes the Earth, v f
= 0.
6-6


Gmm
E r i

1
2 mv i
2 
0 v i

2 Gm
E r i
The starting position is on the Earth’s surface and r i m
E
= 5.97×10
24
kg. The escape velocity is
= R
E
= 6.36×10
6
m. The mass of the Earth is v i

2 Gm r i
E

2 ( 6 .
67

10

11 Nm 2 / kg 2 )( 5
( 6 .
37

10 6 kg )
.
97

10 24 kg )

11 , 200 m/s
This is v i

11 , 200 m s

1 mi
1609 m

7 .
0 mi s
To summarize , we developed what is usually called the work-energy theorem :
W nc
 
K
 
U or K i

U i

W nc

K f

U f where K is the kinetic energy
K

1
2 mv
2 ,
U is the potential energy (usually gravitational)
U grav
 mgy and W nc
is the total work performed by all nonconservative forces. In many situations (but not all), mechanical energy is conserved and W nc
= 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

W

F x
 x
6-7

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
U
 
GM
E m r
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
F x
  kx
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 .
6-8

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
U elastic

1
2 kx
2
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
P av


E
 t
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.
P




E
 t
F
 r
Fv
 t cos
 cos

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,
6-9

0

2

1 kx
1
2 
K
1

U
1 mgy
1



K
2

U
2
1
2 mv
2
2 
2

1 kx
2
2 
0
 v
2

2 gy
1
 k ( x
1
2  x
2
2
) / m

2 ( 9 .
8 m/s

6 .
04 m/s
2
)( 1 .
4 m )

( 28 N/m )(( 0 .
18 m )
2 
( 0 .
12 m )
2
) /( 0 .
056 kg )
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) mg sin



F x
F air
F air
 ma x

0
 mg sin


( 1500 kg )( 9 .
8 m/s
2
) sin 2


513 N
(a) Now the car is moving along a horizontal road at constant velocity. From the FBD found at the top of the next page:
F motor


F
F air x
 ma x

0
F motor

F air
6-10

The power delivered by the engine when the car moves at 20 m/s (notice the angle below is not
2º!) is
P motor

F motor v cos
 
F air v cos 0
 
( 513 N )( 20 m/s ) cos 0
 
10 , 300 W
(b) Again a free body diagram is helpful.


Climbing with constant speed, a x
= 0,
F motor
 mg sin



F x
F air
 ma x

0

F air
 mg sin

F motor
The maximum power supplied by the engine is 40.0 kW. The power is
P



F motor v cos

( F air
 mg sin

) v cos 0
F air v
 mgv sin

6-11

sin
 
P

F air v mgv

40 .
0

10
3
W

( 513 N )( 20 m/s )
( 1500 kg )( 9 .
8 m/s
2
)( 20 m/s )

0 .
101
This corresponds to a 5.8º angle.
6-12