Chapter 3 Energy and Conservation Laws. 2 Conservation laws The most fundamental ideas we have in...
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Transcript of Chapter 3 Energy and Conservation Laws. 2 Conservation laws The most fundamental ideas we have in...
Chapter 3
Energy and Conservation Laws
2
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.”
3
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.
4
Linear momentum
Linear momentum is defined as the product of an object’s mass and its velocity.
We typically just say momentum.
linear momentum mass velocity
p mv
5
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.
6
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 momentumforce
change in time
pF
t
7
Linear momentum, cont’d
How is this related to F = ma?
So, F = ma holds only if the object’s mass remains constant. Not a rocket. Not really a car but its close enough.
mvp vF m ma
t t t
8
ExampleExample 3.1Let’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.
9
ANSWER:
The problem gives us:
The force is:
ExampleExample 3.1
0.06 kg
0 m/s
40 m/s
0.005 s
i
f
m
v
v
t
0.06 kg 40 m/s
0.005 s480 N 108 lb
mvF
t
10
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.
11
Linear momentum, cont’d
It also tells us that we can change the momentum using various forces and time intervals:
Use a large force for a short time, or Use a small force for a long time.
p F t
12
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:
0 0p
F pt
13
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.
14
Conservation of linear momentum, cont’d
We can write this as:
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.
before afterp p
15
ExampleExample 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?
16
ANSWER:
The problem gives us:
The momentum before:
The momentum after:
ExampleExample 3.2
1
2
1,000 kg
1,500 kg
4 m/sf
m
m
v
before 1 1 11,000 kgp m v v
after 1 2 2 2,500 kg 4 m/sp m m v
17
ANSWER:
Conserving momentum
ExampleExample 3.2
1
1
1,000 kg 2,500 kg 4 m/s
2,500 kg4 m/s
1,000 kg
10 m/s
v
v
18
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.
ExampleExample 3.2
19
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.
20
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.”
21
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.
22
Work, cont’d
The force multiplied by the distance moved is the same for both:
input output
3 3F d F d
23
Work, cont’d
We have the same situation for placing a barrels on a loading dock:
2 2
2 2
F d Fd
F d Fd
24
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
25
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).
26
Work, cont’d
From the definition of work:
1 joule (1 newton) (1 meter)
1 J 1 N m
27
ExampleExample 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?
28
ANSWER:
The problem gives us:
The required work is:
ExampleExample 3.3
100 N
3 m
F
d
100 N 3 m
300 J
W Fd
29
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.
30
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.
31
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.
32
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.
33
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.
34
ExampleExample 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?
35
ANSWER:
The problem gives us:
The required work is:
ExampleExample 3.4
30 kg
1.2 m
m
d
230 kg 9.8 m/s 1.2 m
353 J
W Fd mgd
36
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.
ExampleExample 3.4
37
ExampleExample 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?
38
ANSWER:
The problem gives us:
To find work we use:
But we need the distance the car moved.
ExampleExample 3.5
27 m/s
1,000 kg
10 s
2,700 N
v
m
t
F
W Fd
39
ANSWER:
Recall that
The work is
ExampleExample 3.5
2,700 N 135 m
364,500 J
W Fd
2 21 12 2
212
27 m/s10 s 135 m.
10 s
vd at t
t
40
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.
ExampleExample 3.5
41
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.
42
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.
43
Kinetic energy
The formula for kinetic energy is:
m is the object’s mass. v is the object’s speed.
212KE mv
44
ExampleExample 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.
45
ANSWER:
The problem gives us:
The kinetic energy is:
ExampleExample 3.6
27 m/s
1,000 kg
v
m
212
212 1000 kg 27 m/s
364,500 J
KE mv
46
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.
ExampleExample 3.5
47
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 mgPE W Fd mgd
48
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.
49
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.
50
ExampleExample 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.
51
ANSWER:
The problem gives us:
The PE relative to the table is:
ExampleExample 3.7
above table
above floor
3 kg
0.5 m
1.5 m
m
h
h
rel to table above table
23 kg 9.8 m/s 0.5 m
14.7 J
PE mgd mgh
52
ANSWER:
The PE relative to the table is:
ExampleExample 3.7
rel to floor above floor
23 kg 9.8 m/s 1.5 m
44.1 J
PE mgd mgh
53
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.
ExampleExample 3.7
54
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.
55
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.
56
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.
57
Conservation of energy, cont’d
Here are examples of transforming energy:
58
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
59
Conservation of energy, cont’d
To deal with energy conservation, we need the total energy:
total energy:
constantE KE PE
60
Conservation of energy, cont’d
We can use this to solve problems. The initial energy of the ball is all PE:
The final energy is KE:
initial 0E KE PE PE
mgd
final
212
E KE PE
mv
61
Conservation of energy, cont’d
Conserving energy:
Solving for the speed:
initial final
212
E E
mgd mv
2122 2
2
v gd
v gd
62
ExampleExample 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.
63
ANSWER:
The problem gives us:
The initial energy of the man is:
The final energy of the man is:
ExampleExample 3.8
50 mh
i i iE KE PE mgh
212f f fE KE PE mv
64
ANSWER:
Conserve energy:
ExampleExample 3.8
212
212
i fE E
mgh mv
gh v
65
ANSWER:
Solving for the speed gives:
ExampleExample 3.8
2 2 2
2
2 9.8 m/s 50 m 980 m /s
31.3 m/s.
v gh
66
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.
ExampleExample 3.8
67
Conservation of energy, cont’d
Conserving energy:
Or solve for the distance:
initial final
212
E E
mgd mv
212
2
2
vgd
g g
vd
g
68
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.
69
ANSWER:
The problem gives us:
The initial energy of the ball is:
The final energy of the ball is:
Example
0.06 kg
v 20 m/s
m
212i i iE KE PE mv
f f fE KE PE mgh
70
ANSWER:
Conserve energy:
Example
212
212
i fE E
mgh mv
gh v
71
ANSWER:
Solving for the height gives:
Example
22
2
20 m/s
2 2 9.8 m/s
20.4 m.
vh
g
72
DISCUSSION:
Again, since we neglect air resistance the height would be the same if I tossed a tennis ball, bowling ball or brick.
Example
73
Conservation of energy, cont’d
We can understand a roller-coaster by as an example of energy conservation.
74
Collisions
A collision is when two objects interact to: exchange energy, and/or exchange momentum.
75
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.
76
Collisions, cont’d
The first case shows an elastic collision. The second
case shows a perfectly inelastic collision.
77
Collisions, cont’d
In general, the collision does not have to involve physical contact.
78
ExampleExample 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.
79
ANSWER:
The problem gives us:
The initial KE is:
The final KE is:
ExampleExample 3.9
1
1
2
2
1,000 kg
10 m/s
2,500 kg
4 m/s
m
v
m
v
212 1000 kg 10 m/s 50 kJiKE
212 2500 kg 4 m/s 20 kJfKE
80
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.
ExampleExample 3.8
81
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.
82
Power, cont’d
Mathematically,
In SI units, we use the watt: 1 W = 1 J/s.
In English units, we use horsepower: 1 hp = 746 W.
Power: work E
Ptime t
83
ExampleExample 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?
84
ANSWER:
The problem gives us:
The power is:
ExampleExample 3.10
1,000 kg
0 m/s
27 m/s
10 s
i
f
m
v
v
t
364,500 J36,450 W
10 s
EP
t
85
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
ExampleExample 3.10
36,450 W122,000 W
0.30
86
Rotation and angular momentum We defined linear momentum as
Consider an object moving in a circle.
p mv
87
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.
88
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
89
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: i imv r
final: f fmv r
90
Rotation and angular momentum, cont’d
By conservation of momentum:
The right hand side must equal the left hand side.
So as rf gets smaller, vf must get larger.
i i f f
i i f f
mv r mv r
v r v r
91
Rotation and angular momentum, cont’d
This also holds for orbits. As the satellite
gets to B, it must move faster than at A.
92
Example
You spin a ball attached to the end of a 1.0-meter string with a speed of 10 m/s. Find the ball’s speed as you shorten the string to 10 centimeters.
93
ANSWER:
The problem gives us:
From angular momentum conservation:
Example
10 m/s
1 m
0.10 m
i
i
f
v
r
r
ii i f f f i
f
rv r v r v v
r
94
ANSWER:
Inserting the numbers:
Example
1.0 m1 m/s
0.10 m
10 m/s.
if i
f
rv v
r
95
DISCUSSION:
Just a figure skater pulls in her arms, the ball’s speed increases as the string’s length shortens.
Example