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
“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.
“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)
Work


Suppose a force F causes an object to move a distance x parallel to F

F

F
x

The work done by a constant force F is defined as
W  Fx
DO NOT MEMORIZE!



Suppose a constant force F causes an object to move along r not parallel to F

F

F



r
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 0o and 90o
 Negative:  between 90o and 180o
 Zero:  = 90o
o Tension and normal force don’t do work.
The work done by several forces can be found from the net force
Wtotal  W1  W2    WN
 Fnet r cos 
Work and Kinetic Energy
Choosing the x axis along Fnet, (using x = r cos )
Wtotal  Fnet x
 max x
We had an equation from chapter 2
v fx  vix  2ax x
2
2
ax x  12 (v fx  vix )
2
2
Substituting into Wtotal
Wtotal  12 m(v fx  vix )
2
2
Since the net force is in the x-direction, ay and az are both zero. Only the x-component of the
velocity changes
v f  vi  (v fx  v fy  v fz )  (vix  viy  viz )  v fx  vix
2
2
2
2
2
2
2
2
2
2
and
Wtotal  12 m(v f  vi )
2
2
The translational kinetic energy is defined as
K  12 mv2
The work-kinetic energy theorem is
Wtotal  K
While this expression is foundational to this chapter, do not memorize. We shall derive a more
useful form.
Gravitational Potential Energy
The force of gravity 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  Wgrav
In terms of position
U grav  mgy
The gravitational potential energy is
U grav  mgy
The final form of our relation is
Wnc  K  U
This is it. You need to know it. We have another entry into our cause and effect table.
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
E  K U
Conservation of Mechanical Energy
When nonconservative forces do no work, mechanical energy is conserved:
Ei  E f
The zero of potential energy is arbitrary. Choose whatever is convenient.
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.
E1  E2
U1  K1  U 2  K 2
mgy1  12 mv1  mgy2  12 mv2
2
gy1  12 v1  gy2  12 v2
2
2
2
v2  v1  2 g ( y1  y2 )
2
 (5 m/s ) 2  2(9.8 m/s 2 (10 m  5 m)
 11.1 m/s
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.
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.
Ei  E f
U i  Ki  U f  K f

GmmE 1
GmmE 1
2
2
 2 mvi  
 2 mv f
ri
rf
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.

GmmE 1
2
 2 mvi  0
ri
vi 
2GmE
ri
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
vi 
2GmE
2(6.67  1011 Nm2 / kg 2 )(5.97  1024 kg)

 11,200 m/s
ri
(6.37  106 kg)
This is
vi  11,200
m
1 mi
mi

 7.0
s 1609 m
s
To summarize, we developed what is normally called the work-energy theorem:
Wnc  K  U
where K is the kinetic energy
K  12 mv2 ,
U is the potential energy (usually gravitational)
U grav  mgy
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 to the right.
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 is
W  Fx x
Fig. 06.26
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
Fx  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.
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  12 kx2
Note that U = 0 when x = 0.
Power
Sometimes the rate of energy conversion is important. We use the term power to refer to the rate
of energy conversion.
P
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.
E
t
Fr cos 

t
 Fv cos 
P
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 a 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,
K1  U1  K 2  U 2
0  12 kx1  mgy1  12 mv2  12 kx2  0
2
2
2
v2  2 gy1  k ( x1  x2 ) / m
2
2
 2(9.8 m/s 2 )(1.4 m)  (28 N/m)((0.18 m) 2  (0.12 m) 2 ) /( 0.056 kg)
 6.04 m/s
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 weight of the car is opposed by air resistance. The free body
diagram is
N
Fair

mg

(a) For the x-component (along the incline)
F
x
 max
mg sin   Fair  0
Fair  mg sin 
 (1500 kg)(9.8 m/s 2 ) sin 2
 513 N
The power delivered by the engine when the car moves at 20 m/s (notice the angle below is not
2º!)
P  Fv cos   Fairv cos 0  (513 N)(20 m/s ) cos 0  10,300 W
(b)
Again a free body diagram is helpful.
N
Fmotor
Fair
mg 

Climbing with constant speed,
F
x
 max
Fmotor  mg sin   Fair  0
Fmotor  Fair  mg sin 
The maximum power supplied by the engine is 40.0 kW. The power is
P  Fmotorv cos 
 ( Fair  mg sin  )v cos 0
 Fairv  mgv sin 
sin  

P  Fairv
mgv
40.0 103 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.