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Chapter 3
Energy and Conservation Laws
Conservation laws
 The most fundamental ideas we have in
physics are conservation laws.

Statements telling us that some quantity
does not change.
 Conservation of mass states:
 The total mass of an isolated system is
constant.
 To apply these, we must define a “system.”
2
Conservation laws, cont’d
 A system is just a collection of objects we
decide to treat at one time.


The tanker and fighter can represent a
system.
The fuel leaving
the tanker goes
into the fighter:
mass is conserved.
3
Linear momentum
 Linear momentum is defined as the product
of an object’s mass and its velocity.
linear momentum  mass  velocity
p  mv

We typically just say momentum.
4
Linear momentum, cont’d
 Momentum is a measure of an object’s state
of motion.

Consider an object whose momentum is
1 kg·m/s


This could be a 0.005 kg bullet traveling at 200
m/s.
This could be a 0.06 kg tennis ball traveling at
16.7 m/s.
5
Linear momentum, cont’d
 Newton’s 2nd law is closely related to
momentum.

The net external force acting on an object
equals the rate of change of linear momentum:
change in momentum
force 
change in time
p
F
t
6
Linear momentum, cont’d
 How is this related to F = ma?
p   mv 
v
F

m
 ma
t
t
t
 So, F = ma holds only if the object’s mass
remains constant.


Not a rocket.
Not really a car but its close enough.
7
Example
Example 3.1
Let’s estimate the average force
on a tennis ball as it is served.
The ball’s mass is 0.06 kg and
it leaves the racquet with a
speed of 40 m/s. High-speed
photography indicates that the
contact time is about
5 milliseconds.
8
Example
Example 3.1
ANSWER:
The problem gives us:
m  0.06 kg
vi  0 m/s
v f  40 m/s
The force is:
F
  mv 
t  0.005 s
0.06 kg  40 m/s 


t
0.005 s
 480 N  108 lb
9
Linear momentum, cont’d
 This tells why we must exert a force to stop
an object or get it to move.


To stop a moving object, we have to bring its
momentum to zero.
To start moving an object, we have to impart
some momentum to it.
10
Linear momentum, cont’d
 It also tells us that we can change the
momentum using various forces and time
intervals:
p  F t


Use a large force for a short time, or
Use a small force for a long time.
11
Conservation of linear
momentum
 The Law of Conservation of Linear
Momentum states:
The total linear momentum of an isolated
system is constant.
 Isolated implies no external force:
p
F 0
 p  0
t
12
Conservation of linear
momentum, cont’d
 This law helps us deal with collisions.
 If the system’s momentum can not change,
the momentum before the collision must
equal that after the collision.
13
Conservation of linear
momentum, cont’d
 We can write this as:
pbefore  pafter
 To study a collision:
 Add the momenta of the objects before the
collision.
 Add the momenta after the collision.
 the two sums must be equal.
14
Example
Example 3.2
A 1,000 kg car (car 1) runs into the rear of a stopped
car (car 2) that has a mass of 1,500 kg. Immediately
after the collision, the cars are hooked together and
have a speed of 4 m/s. What was the speed of car 1
just before the collision?
15
Example
Example 3.2
ANSWER:
The problem gives us:
The momentum before:
m1  1, 000 kg
m2  1,500 kg
v f  4 m/s
pbefore  m1v1  1,000 kg  v1
The momentum after:
pafter   m1  m2  v2   2,500 kg  4 m/s 
16
Example
Example 3.2
ANSWER:
Conserving momentum
1, 000 kg  v1   2,500 kg  4 m/s 
2,500 kg
v1 
 4 m/s 
1,000 kg
 10 m/s
17
Example
Example 3.2
DISCUSSION:
Both cars together have more mass than just
car 1.
Since both move away at 4 m/s, the lighter car
1 must have a greater speed before the
collision.
18
Conservation of linear
momentum, cont’d
 How do rockets work?

The exhaust exits the rocket at high speed.


We need high speed because the gas has little
mass.
The rocket moves in the opposite direction.

Not as fast as the
gas b/c more mass.
19
Work
 Imagine using a lever to lift a heavy object.
 The lever allows us to exert less force than
the object
actually weighs.
 This sounds like
“free money.”
20
Work, cont’d
 There’s a catch:

We have to apply our force through a greater
distance than the rock moves.
 So there must be
some connection
between force
and distance.
21
Work, cont’d
 The force multiplied by the distance moved is
the same for both:
input  output
 F  3d    3F  d 
22
Work, cont’d
 We have the same situation for placing a
barrels on a loading dock:
 2 F  d   2 Fd
 F  2d   2 Fd
23
Work, cont’d
 Work is defined as the product of force and
the distance through which the force moves
an object in the direction of the force.
work  w  Fd
24
Work, cont’d
 The units of work:

Metric





SI: joule (J = N·m),
erg (= 10-7 J),
calorie (cal = 4.186 J),
kilowatt-hour (kWh = 0.278 J).
English:


foot-pound (ft·lb),
British thermal unit (Btu).
25
Work, cont’d
 From the definition of work:
1 joule  (1 newton)  (1 meter)
1 J 1 Nm
26
Example
Example 3.3
Because of friction, a constant force of
100 newtons is needed to slide a box across a
room. If the box moves 3 meters, how much must
be done?
27
Example
Example 3.3
ANSWER:
The problem gives us:
F  100 N
d 3m
The required work is:
W  Fd  100 N  3 m 
 300 J
28
Work, cont’d
 Recall that force is a vector.

Involves magnitude and direction.
 Work is just that part of the force in the
direction of the displacement.

Work is not a vector — it’s a scalar.
 But the sign of the work does depend on the
relative directions.
29
Work, cont’d
 If the force and distance are in the same
direction, the force does positive work.
 If the force and distance are in the opposite
direction, the force does negative work.
30
Work, cont’d
 If the force is not in the direction of the
direction, the force does no work.



The string’s tension is toward the center of
the circle.
The ball moves along
the circle’s
circumference.
So, the tension does
no work.
31
Work, cont’d
 You do positive work (in the physics-sense) when
you lift the create.
 You do NO work (in the
physics-sense) when you
carry the crate.
 You do negative work
when you set the crate
down.
32
Work, cont’d
 When you throw or catch a ball, you do work
on the ball.


Your hand exerts a force on the ball.
You exert that force through the throwing or
catching distance.

If you’re strong, you
don’t need the same
distance because of
the larger force.
33
Example
Example 3.4
Let’s say that the barrel has a mass of 30 kg and that
the height of the dock is 1.2 meters. How much work
would you do when lifting the barrel?
34
Example
Example 3.4
ANSWER:
The problem gives us:
m  30 kg
d  1.2 m
The required work is:
W  Fd  mgd

  30 kg  9.8 m/s
 353 J
2
 1.2 m 
35
Example
Example 3.4
DISCUSSION:
You would do the same amount of work rolling
the barrel up the ramp.
You would only have to exert a force of 150 N
instead of the entire 300 N.
But you have to exert that smaller force over a
distance of 2.4 m.
36
Example
Example 3.5
In Example 2.2 we used Newton’s 2nd law to compute
the force needed to accelerate a 1,000-kg car from
0 to 27 m/s in 10 seconds. The answer was 2,700
N. How much work is done?
37
Example
Example 3.5
ANSWER:
The problem gives us:
v  27 m/s
m  1, 000 kg
t  10 s
F  2, 700 N
To find work we use:
W  Fd
But we need the distance the car moved.
38
Example
Example 3.5
ANSWER:
Recall that
The work is
v 2
d  at 
t
t
2
1 27 m/s
2
10 s   135 m.
10 s
1
2
2
1
2
W  Fd   2, 700 N 135 m 
 364,500 J
39
Example
Example 3.5
DISCUSSION:
In reality, this is smaller than the energy the
engine must generate.
The engine must overcome its internal friction
— a loss of energy.
Most cars are about 30% efficient.
So you need (364 kJ)/(0.3) = 1.2 MJ to actually
accelerate this car.
40
Energy
 Energy is defined as the measure of a
system’s ability to do work.


We use the symbol E to represent energy.
Energy has the same units as work:

Joule for SI, ft·lb for English.
41
Energy, cont’d
 There are various types of energy.

Kinetic energy is the energy associated with
an object’s motion.


We use the symbol KE.
Potential energy is energy associated with
the system’s position or orientation.

We use the symbol PE.
42
Kinetic energy
 The formula for kinetic energy is:
KE  mv
1
2
 m is the object’s mass.
 v is the object’s speed.
2
43
Example
Example 3.6
In Example 3.5 we computed the work that is done on
a 1,000-kg car as it accelerates from 0 to 27 m/s.
Find the car’s kinetic energy when it is traveling at
27 m/s.
44
Example
Example 3.6
ANSWER:
The problem gives us:
v  27 m/s
m  1, 000 kg
The kinetic energy is:
KE  mv
1
2

1
2
2
1000 kg  27 m/s 
2
 364,500 J
45
Example
Example 3.5
DISCUSSION:
This equals the (ideal) work required to get the
car up to speed.
We could determine how much work is
required by finding the kinetic energy of the
car.
This is the idea of energy conservation.
46
Gravitational potential energy
 Gravitational potential energy equals the
work done by the gravity.

If you lift an object, you must apply a force at
least equal to the object’s weight:

Lifting it through a distance d, the work is
F  mg
PE  W  Fd  mgd
47
Gravitational potential energy,
cont’d
 Note that we only deal with the distance
through which the object moves.


The brick has 14.7 J
of PE relative to the
table top.
It has 44.1 J of PE
relative to the floor.
48
Gravitational potential energy,
cont’d
 Where you say an object has zero PE is
arbitrary.

We only care about the change.


Let’s say the ground
is at zero PE.
In the hole, the ball
has negative PE.
 It is below the
reference level.
49
Example
Example 3.7
A 3-kg brick is lifted to a height of 0.5 meters above a
table that is 1.0-m tall. Find the gravitational
potential energy relative to the table and the floor.
50
Example
Example 3.7
ANSWER:
The problem gives us:
m  3 kg
habove table  0.5 m
The PE relative to the
table is:
habove floor  1.5 m
PErel to table  mgd  mghabove table

  3 kg  9.8 m/s
 14.7 J
2
  0.5 m 
51
Example
Example 3.7
ANSWER:
The PE relative to the table is:
PErel to floor  mgd  mghabove floor

  3 kg  9.8 m/s
2
 1.5 m 
 44.1 J
52
Example
Example 3.7
DISCUSSION:
The PE is meaningless without specifying the
reference level.
If we drop the brick:
It has a lower speed upon impact with the table than it
does with the floor.
The brick has more energy to convert to KE relative to
the floor.
53
Internal energy
 Internal energy is that energy associated
with the internal structure of the object.


A hot rock has more internal energy than a
cold rock.
Lifting either increases the external PE but not
the internal energy.
54
Internal energy, cont’d
 Internal energy can be used to do work.


If you turn on an electric stove, the internal
energy of the filament increases.
This energy can be used to boil water.


You increase the internal energy of the water.
The produced steam can be used to power a
turbine/generator.
55
Conservation of energy
 The Law of Conservation of Energy:
energy cannot be created or destroyed.

The total energy of an isolated system is
constant.

The energy of the Universe is constant.
 Energy can only be transformed from one
form to another.
56
Conservation of energy, cont’d
 Here are examples of transforming energy:
57
Conservation of energy, cont’d
 If the energy of an isolated system is
constant, the energy before an event must
the same as the energy after an event.
total energy before  total energy after
58
Conservation of energy, cont’d
 To deal with energy conservation, we need
the total energy:
total energy:
E  KE  PE  constant
59
Conservation of energy, cont’d
 We can use this to solve problems.

The initial energy of the ball is all PE:
Einitial  KE  PE  0  PE
 mgd

The final energy is KE:
Efinal  KE  PE
 mv
1
2
2
60
Conservation of energy, cont’d
 Conserving energy:
Einitial  Efinal
mgd  mv
1
2
2
 Solving for the speed:
2  v  2  gd
1
2
2
v  2 gd
61
Example
Example 3.8
In 2003, a man went over Horseshoe Falls, part of
Niagara Falls, and survived. The height of the falls is
about 50 meters. Estimate the speed of the man
when he hit the water at the bottom of the falls.
62
Example
Example 3.8
ANSWER:
The problem gives us:
h  50 m
The initial energy of the man is:
Ei  KEi  PEi  mgh
The final energy of the man is:
E f  KE f  PE f  mv
1
2
2
63
Example
Example 3.8
ANSWER:
Conserve energy:
Ei  E f
mgh  mv
1
2
gh  v
1
2
2
2
64
Example
Example 3.8
ANSWER:
Solving for the speed gives:
v  2 gh

 2 9.8 m/s
2
 50 m  
2
980 m /s
2
 31.3 m/s.
65
Example
Example 3.8
DISCUSSION:
The speed does not depend on the man’s
mass.
If you tried this, you’d hit the bottom with the
same speed.
This is obviously ideal:
We do not consider air resistance. That would convert some
of his KE into heat and sound. The real speed would be
slower.
66
Conservation of energy, cont’d
 Conserving energy:
Einitial  Efinal
mgd  mv
1
2
2
 Or solve for the distance:
1
2
v
gd

g
g
v2
d
2g
2
67
Example
I toss a 0.06-kg tennis ball straight up. When it leaves
my hand, it has a speed of 20 m/s. Find how high
the ball rises.
68
Example
ANSWER:
The problem gives us:
m  0.06 kg
v  20 m/s
The initial energy of the ball is:
Ei  KEi  PEi  mv
1
2
2
The final energy of the ball is:
E f  KE f  PE f  mgh
69
Example
ANSWER:
Conserve energy:
Ei  E f
mgh  mv
1
2
gh  v
1
2
2
2
70
Example
ANSWER:
Solving for the height gives:
20 m/s 

v
h

2 g 2 9.8 m/s 2
2
2


 20.4 m.
71
Example
DISCUSSION:
Again, since we neglect air resistance the
height would be the same if I tossed a
tennis ball, bowling ball or brick.
72
Conservation of energy, cont’d
 We can understand a roller-coaster by as an
example of energy conservation.
73
Collisions
 A collision is when two objects interact to:


exchange energy, and/or
exchange momentum.
74
Collisions, cont’d
 An elastic collision is one in which the total
kinetic energy remains constant.
 An inelastic collision is one in which the
total kinetic energy does not remain constant.
75
Collisions, cont’d
 The first case shows an elastic collision.
 The second
case shows a
perfectly
inelastic
collision.
76
Collisions, cont’d
 In general,
the collision
does not
have to
involve
physical
contact.
77
Example
Example 3.9
Recall the automobile collision analyzed in Example
3.2. Compare the amounts of kinetic energy in the
system before and after the collision.
78
Example
Example 3.9
ANSWER:
The problem gives us:
m1  1, 000 kg
v1  10 m/s
m2  2,500 kg
v2  4 m/s
The initial KE is:
KEi 
1
2
1000 kg 10 m/s 
2
 50 kJ
2
 20 kJ
The final KE is:
KE f 
1
2
 2500 kg  4 m/s 
79
Example
Example 3.8
DISCUSSION:
30 kJ of energy was lost during the collision.
Converted to heat and sound, used to deform the cars,
etc.
Since KE was lost, the collision was inelastic.
It is perfectly inelastic because they stuck together.
80
Power
 Energy tells you about the position and
speed.
 But it contains no information about time.
 Power is the rate at which energy is
transferred or transformed.

The rate of doing work.
81
Power, cont’d
 Mathematically,
work E
Power: P 

time
t

In SI units, we use the watt:


1 W = 1 J/s.
In English units, we use horsepower:

1 hp = 746 W.
82
Example
Example 3.10
In Examples 2.2 and 3.5, we computed the
acceleration, force and work for a 1,000-kg car that
goes from 0 to 27 m/s in 10 s. We can now
determine the required power output of the engine.
What is its value?
83
Example
Example 3.10
ANSWER:
The problem gives us:
m  1, 000 kg
vi  0 m/s
v f  27 m/s
t  10 s
The power is:
E 364,500 J
P 
 36, 450 W
t
10 s
84
Example
Example 3.10
DISCUSSION:
This is the ideal value.
For a 30% efficient car to reach 27 m/s, to
overcome the losses due to friction the
engine must generate a power of
36, 450 W
 122, 000 W
0.30
85
Rotation and angular
momentum
 We defined linear momentum as
 Consider an
p  mv
object
moving in a
circle.
86
Rotation and angular
momentum, cont’d
 If we shorten the string, the object’s speed
increases.

There must
be some
relation
between the
speed and
radius.
87
Rotation and angular
momentum, cont’d
 Angular momentum is the momentum of an
object moving in a circle.
 For a point particle, e.g., a ball, the angular
momentum is
mvr
88
Rotation and angular
momentum, cont’d
 Conservation of Angular Momentum states
that the total angular momentum of an
isolated system is constant.

For our ball on a string:
initial: mvi ri
final: mv f rf
89
Rotation and angular
momentum, cont’d
 By conservation of momentum:
mvi ri  mv f rf
vi ri  v f rf


The right hand side must equal the left hand
side.
So as rf gets smaller, vf must get larger.
90
Rotation and angular
momentum, cont’d
 This also holds for orbits.

As the satellite
gets to B, it
must move
faster than at
A.
91
Example
You spin a ball attached to the end of a 1.0meter string with a speed of
10 m/s. Find the ball’s speed as you shorten
the string to 10 centimeters.
92
Example
ANSWER:
The problem gives us:
vi  10 m/s
ri  1 m
rf  0.10 m
From angular momentum conservation:
vi ri  v f rf
ri
 v f  vi
rf
93
Example
ANSWER:
Inserting the numbers:
ri
1.0 m
v f  vi 
1 m/s 
rf
0.10 m
 10 m/s.
94
Example
DISCUSSION:
Just a figure skater pulls in her arms, the ball’s
speed increases as the string’s length
shortens.
95