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Chapter 3
Energy (Spring 09)
What 4-letter word in the English
language upsets more people than
any other?

Especially college students
Work

Work is defined as being the product of a force
and the distance the force moves in the
direction of the force.
–
–
–
–
–
Movement must happen in the direction of the force.
The force and the motion cannot be perpendicular.
No work is done in holding something.
No work is done in carrying something parallel to
the ground.
No work is done in doing physical science
homework unless a pencil or pen is pushed along
the paper.
Work
Work = force x distance moved
W = F • d (put on formula sheet)
What work is done when a force of 10 N is used to push
an object 3 m?
Data
W=?
F=10N
d=3m
Formula
W=F•d
Work
W=10N•3m=
Answer
30 joules
Note that the weight or mass of the object is not
important, only the force necessary to move it.
The joule
1 joule = 1 newton • 1 meter
A joule is the work done in pushing with a force
of one newton for a distance of one meter.
James Prescott Joule was an English brewmaster. We will hear more of him later.
3q
Work against gravity


Work = force x distance
When we lift something, the force necessary to
lift it is the weight of the object
–

w=mg (force)
Therefore, the work necessary to raise an object
a height h is
–
–
W=mgh (formula sheet) = work
h is the height the object is raised (or lowered).

The weight is w=mg, don’t confuse weight and work. Weight
is a force that can do work.
Calculate the amount of work for a 100 kg
person to climb stairs that are 10 m high.
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Data
W=?
m=100 kg
h=10 m
W  F  d  mgh
W  100kg  9.8m / s 10m  9800 j
2
How fast can you work?
Power is the rate of doing work.
 Power = work/time
W
P
Formula sheet!
t
If 30 joules of work is done in 6 seconds, what is the power?

Data
Formula
Work
Answer
W=30 j
t=6 s
P=?
P = W/t
P = 30j/6s
5?
P
W
t
Units of power

One watt is one joule per second.

Watt = joule/sec

5 watts = 5 w
What is my power if my mass is 70 kg
and I can climb stairs that are 10 m high
in 3 seconds?
Data
P=?
m=70kg
h=10m
t=3 s
Formula Work
Answer
P = W/t P = W/t=F d/t= mgh/t
2286 W
W=F·d P=70kg9.8m/s210m/3s
F=mg
d=h
One horsepower is 746 watts so 2286 W is 3.1 horsepower.
This calculation is necessary for your pre-lab next week
(measure your horsepower).
Note that this says nothing about how steep the stairs
are, only their height.
2q
If I have a lot of energy I can do a
lot of work.


Energy is the ability (capacity) to do work.
Energy comes in many forms:
–
–
Heat energy
Chemical energy



–
–
Food
Gasoline
Dynamite
Potential energy
Kinetic energy
Potential energy


Potential energy is energy due to the position
of an object.
Examples of potential energy include:
–
–
–
A compressed spring
A nail near a magnet
A car on a hill

Gravitational potential energy
Gravitational Potential Energy





Energy = work done
To have one joule of energy, I must do 1 joule
of work to raise the object.
The units of work are also joules.
The force of gravity is mass times the
acceleration of gravity
W = mg
Gravitational P.E. (cont)




P.E. = Work = force • distance
W=f•d
f = Weight = W = mg
d = distance = height = h
P.E. = mgh (Formula for gravitational potential
energy)
This applies only on the surface of the Earth.
Gravitational P.E. (cont)
What is the P.E. of a 20 kg block, 2 m above the
floor?
Data
Formula
Work
Answer
P.E. = ?
m = 20 kg
h=2m
g = 9.8m/s2
P.E. = mgh P = 20kg• 9.8m/s2•2m 392 j
Kinetic Energy

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
Kinetic energy is the energy something has
because it is moving.
Energy of motion
K.E. = ½ mv2
Where m = mass of the object
v = velocity of the object
Kinetic energy
What is the K.E. of a 70 kg person running 10 m/s?
Data
Formula
Work
K.E. = ?
m = 70 kg
V= 10 m/s
K.E. = ½ mv2 = ½ • 70kg•(10m/s)2
Answer
3500 j
Energy – a review



Energy – the ability to do work
Potential energy P.E. = mgh
Kinetic energy K.E. = ½ mv2
Conservation of Energy
Energy may be changed
from one type to another
but the total amount of
energy remains the
same.
More on conservation of energy


Total energy is always conserved
We will be especially interested in situations in
which mechanical energy is conserved
–



This means no energy is given to heat or friction
When mechanical energy is conserved, the sum
of the potential and the kinetic energies will
remain constant.
There are many types of problems we can solve
using this knowledge.
This will be the idea that will be worth more
points on the next two tests than any other topic.
Energy conservation- Changing P.E. to K.E. and K.E. to
P.E. - - - - Case 1: A falling object

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
100% P.E
90% P.E.
80% P.E.
70% P.E.
60% P.E.
50% P.E.
40% P.E.
30% P.E.
20% P.E.
10% P.E.
.
at the top
10%K.E.
20%K.E
30%K.E
40%K.E
50%K.E
60%K.E
70%K.E
80%K.E
90%K.E
100%K.E
half way down
at the bottom
P.E. to K.E. and K.E. to P.E.





At the top – All P.E.
1/3 of the way from the top 2/3 P.E., 1/3 K.E.
Half way down ½ K.E., ½ P.E.
1/3 of the way from the bottom, 1/3 P.E., 2/3 K.E.
At the bottom – All K.E.
You will be asked to add one additional
step in the solution of these problems.
That step is the principle.
In the principle you must give the state of
the energy of the system in its initial state
and its final state.
For this problem the principle is: Potential
Energy is changed to Kinetic Energy.
mgh = ½
2
mv
Potential to Kinetic

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What is the velocity of an object after falling 10m?
v=?
3q
h= 10m
Initial state = Potential energy
Final state = kinetic energy
Principle
P.E. = K.E.
mgh = ½ mv2
v  2 g h
v  2  g  h  2  9.8m / s  10m  14m / s
2
Kinetic to Potential



How high will an arrow go that is shot straight into the
air with an initial velocity of 100 m/s?
h = ?, v = 100 m/s 3q
Initial state = kinetic energy
–
–



In this problem the arrow has already been shot and is
moving with a velocity of 100 m/s.
But it is still at zero height.
2
v
h
2 g
final state = potential energy
½ mv2=mgh
Solve the equation for h
2
2
v
(100m / s)
h

?
2
2  g 2  9.8m / s
h  510m
The energy can be used to do work or work
can be used to give energy.


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Work to Kinetic energy
Work to Potential energy
Kinetic energy to work
Potential energy to work
Work to energy and energy to work

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


A 3 kg hammer moving 10 m/s hits a nail and drives it 2
cm into a board. What was the force on the nail?
m = 3 kg v = 10 m/s d = 2 cm = 0.02 m F=? 3q
A falling hammer (K.E.) drives a nail (work)(force x
distance)
2
m

v
Principle K.E. = work
F
2
½ mv =F·d
2d
F
3kg  (10m / s)

?
2  0.02m
2
mv 2
2d
F  7500 N
How fast will a 5 kg rocket be traveling if
its motor has a force of 2000 n and it
operates for 500 m horizontally?
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v=?, m=5kg, F=2000N, d=500m
Principle: work = K.E.
F·d=½ mv2
v=?
m=5kg
F = 2000 n
d = 500 m
2  2000n  500m
v
?
5kg
v
2 Fd
m
v  632m / s
Work to energy and energy to work





A 3 kg hammer moving 10 m/s hits a nail and drives it 2
cm. What was the force on the nail?
m = 3 kg v = 10 m/s d = 2 cm = 0.02 m F=?
A falling hammer (K.E.) drives a nail (work)(force x
distance)
2
m

v
Principle K.E. = work
F
2
½ mv =F·d
2d
F
3kg  (10m / s)

?
2  0.02m
2
mv 2
2d
F  7500 N
How fast will a 5 kg rocket be traveling if
its motor has a force of 2000 n and it
operates for 500 m?

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



v=?, m=5kg, F=2000N, d=500m
Principle: work = K.E.
F·d=½ mv2
v=?
m=5kg
F = 2000 n
d = 500 m
2  2000n  500m
v
?
5kg
v
2 Fd
m
v  632m / s
Conservation of momentum
Two types: Linear momentum and angular
momentum
Linear momentum = mass times velocity
In the absence of outside forces, the total
momentum of a group of objects remains
unchanged (even if the objects collide).
Useful when objects collide or when they separate
(boy jumping from wagon, etc)
Angular momentum = radius x mass x velocity
Useful for systems, such as planets, comets, etc
We will only do calculations with linear momentum.
momentum before collision = momentum after
m1
v1i
v2i
m1 is the mass of object one.
v1i is the initial velocity of object one.
m2 is the mass of object two.
v2i is the initial velocity of object two.
v1f is the final velocity of object one.
v2f is the final velocity of object two.
m1 v1i + m2 v2i = m1 v1f + m2 v2f
m2
This needs to
go on your
formula sheet!
Example one
A 500 kg car traveling 20 m/s collides head-on
with a 2000 kg truck traveling 10 m/s in the
opposite direction. The two stick together after
the collision. What is the direction and velocity
of the two after the collision?
This
minus
sign is
very
important
DATA
m1= 500 kg
v1i = 20 m/s
m2= 2000 kg
v2i = -10 m/s
v1f = v2f = vf = ?
General equation
m1 v1i + m2 v2i = m1 v1f + m2 v2f
In our special case (v1f = v2f = vf )
m1 v1i + m2 v2i = (m1 + m2)vf
500 kg•20 m/s+ 2000 kg•(-10 m/s)=(500+2000)kg•vf
10,000 kg m/s-20,000 kg m/s = 2500 kg • vf
-10,000 kg m/s = 2500 kg • vf
vf = -10,000 kg m/s
2500 kg
vf = -4 m/s Direction is that of the truck
Another momentum example

A 70 kg man stands on a 20 kg boat in the
water and jumps with a velocity of 5 m/s.
What is the recoil velocity of the boat?
m1= 70 kg
v1i = 0
v1f = 5 m/s
m2= 20 kg
v2i = 0
v2f = ?
General equation
m1 v1i + m2 v2i = m1 v1f + m2 v2f
In our special case (v1i = v2i = 0 )
0 =m1 v1f + m2 v2f
- m1 v1f = m2 v2f
The minus sign
tells us the
direction.
-70 kg•5 m/s = 20 kg•vf
vf = -350 kg m/s
20 kg
vf = -17.5 m/s Direction is
opposite the way the man jumped.
Caloric



A colorless, odorless liquid that is in most
substances.
When a substance burns the caloric in it is
given off as heat
This theory explained heat until the 18th
century.
Benjamin Thompson
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An American who fled the U.S. in 1773
because of British sympathies.
Became a military consultant (mercenary)
Took the name of Count Rumford
Made cannon
Discovered that heat is a form of energy
Set up the first public school for the children
of his workers.
James Prescott Joule




An rich English beer maker who liked to do
science experiments.
James Prescott Joule showed how many
joules (unit of energy) were in a calorie (unit
of heat).
1 calorie = 4.2 joules (put on formula sheet)
1 calorie is the amount of heat necessary to
raise the temperature of one gram of water
one degree Celsius.
Guess who
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Didn’t like grade school very well
Got a doctor’s degree in physics but couldn’t get a job
teaching (even in high school) so took a job in the post
office (taking care of patents).
Takes his girl friend on a trip and she gets pregnant.
Child is taken by girl’s relatives, and is not heard of later
Writes three papers in physics journals in 1905 that
profoundly change three major areas of physics.
Becomes most famous scientist of 20th century
Albert Einstein
Theory of Brownian motion
 Theory of photons
 Theory of relativity
Two postulates of special relativity
1) The speed of light is always measured to be
the same even when two systems are
moving with respect to one another.
2) The laws of physics apply in systems that
are moving with respect to one another.

General Theory of Relativity


The special theory that we have already
discussed does not apply to systems that are
accelerated or in a gravitational field.
The general theory applies to accelerated
systems or systems in a gravitational field.
–
The most significant result of the general theory we
will note is that gravity affects light.

Light is bent by gravity (stars)
–
We can see this when there is an eclipse of the sun.
E = mc2
E = Energy
m = mass
c = speed of light = 3 x 108 m/s (formula sheet)
c2 = 9 x 1016 m2/s2
What energy is obtained when 7 grams is
changed to mass
Data: m = 7 grams = 0.007 kg
Data
m = .007kg
E=?
Formula
E=mc2=(.007 kg)•(3 x 108 m/s)2
= 6.3 x 10 14 joules
The Energy Problem
Energy is necessary
for our life style.
We are running out of
the energy sources
we use the most.
Fossil Fuels

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


The cheapest
The easiest to use
The most convenient
The worst for the environment (Global
warming and acid rain)
Supplies will become limited in our lifetime.
Energy Costs per kilowatt hour of
electricity



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
Natural gas
Coal
Wind
Geothermal
Hydropower
3-4¢
4-5¢
4-5¢
3-8¢
4-10¢

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

Biomass
Nuclear
Solar thermal
Photovoltaics
6-8¢
10-15¢
10-15 ¢
20-30¢
Type
Initial
cost
Operating Environmental cost
cost
Coal
Oil
Natural gas
Wind
Solar energy
Hydro power
Low
Low
Low
High
High
High
Lowest
Low
Low
Low
Low
Low
Tide energy
Geothermal
Ocean
thermal
High
High
High
High
Experi- Unknown
mental
CO2. Acid rain
CO2
CO2 ,best ff for environ.
Noise, kills birds
Occupies space
Needs a dam, rainfall
Very bad for fish, etc.
Few places where possible
unknown
Solar generated hydrogen

Burns without pollution
–


H2 + O2  H2O
Could be used in cars and distributed at filling
stations
Car manufacturers are designing hydrogen cars
Cars without engines
Each wheel has an
electric motor that
drives it
Skateboard concept
All electronic
steering, brakes
Tops plug-in to the
fuel cell base.
2q
Best possibilities for the future

Conservation
–
–

Methane hydrate
–
–
–

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
Energy efficient cars/trucks
Better insulated houses
Methane (natural gas) combined with water
Uncertain quantities
In difficult places to access
Increased use of wind and solar energies
Hydrogen powered cars/trucks
Fusion
Nuclear energy


Good or Bad?
Two types
–
Fission (to split apart) (Atomic bomb)



–
Uses Uranium as a fuel
What we now have
All present nuclear power plants use fission
Fusion (to join together) (Hydrogen bomb)


A possibility for the future
Uses hydrogen as a fuel (sea water)
Fission
Fuel is uranium (which splits apart)
Good points
 No CO2
 No radioactivity in normal operation
 In normal operation a nuclear power plant releases much
less radioactivity than a coal burning plant.
Bad points
 Possibility of an accident (Chernobyl)
 Disposing of spent (used) fuel

Fusion
Fuel is hydrogen (not radio-active)
Good points
 Produces tremendous amounts of energy
 Very good for the environment
 Uses ocean water for fuel
Bad points
 We don’t know how to do it yet.

Conservation of energy revisited

This will be the problem on the chapter 3 and
chapter 4 tests that will be worth the most points.
Energy conservation- Changing P.E. to K.E. and K.E. to
P.E. - - - - Case 1: A falling object

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


100% P.E
90% P.E.
80% P.E.
70% P.E.
60% P.E.
50% P.E.
40% P.E.
30% P.E.
20% P.E.
10% P.E.
.
at the top
10%K.E.
20%K.E
30%K.E
40%K.E
50%K.E
60%K.E
70%K.E
80%K.E
90%K.E
100%K.E
half way down
at the bottom
P.E. to K.E. and K.E. to P.E.

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

At the top – All P.E.
1/3 of the way from the top 2/3 P.E., 1/3 K.E.
Half way down ½ K.E., ½ P.E.
1/3 of the way from the bottom, 1/3 P.E., 2/3 K.E.
At the bottom – All K.E.
It is very important to note that the total energy is
always the same
 Total Energy = Potential Energy + Kinetic Energy
(True when there is no friction and no work done).
In the end, the energy will be all heat energy!

Calculate the work necessary to raise an
10 kg rock 30 m high. (25 point problem)

Data
m = 10 kg, h = 30 m, Work=?
W=F·d=mgh=10kg·9.8m/s2·30m = 2940j

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What is the potential energy at this height
(30m)?
No calculation is necessary, the P.E. is 2940J.
What is the kinetic energy at this height?
No calculation is necessary, as the rock is sitting
still, the K.E. is zero!
Total energy = K.E. + P.E. = 2940j at all heights.





What is the total energy at this height?
No calculation is necessary, T.E. is 2940j.
Calculate the P.E., the K.E. and the T.E when
the rock is 10 m from the ground and when it just
begins to touch the ground (the height is zero
and the velocity is maximum).
The total energy remains the same in all cases,
2940j,
At ground level, h=0 so P.E. is zero, thus the
K.E. is 2940j
To calculate the P.E. at 10 m height,
Data: m=10 kg, h=10m, P.E.=?
P.E. = mgh = 10kg·9.8m/s2·10m = 980j
The K.E. at this point is:
K.E. = T.E. – P.E. = 2940j – 980j = 1960j
What is the heat energy produced when the rock
hits the ground?
No calculation is necessary, all the energy is
changed to heat, so the answer is 2940j.

The energy can be used to do work or work
can be used to give energy. – The Principle

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
Work to Kinetic energy
Work to Potential energy
Kinetic energy to work
Potential energy to work
Potential energy to Kinetic energy
Kinetic energy to Potential energy
A 3 kg hammer moving 10 m/s hits a nail
and drives it 2 cm. What was the force
on the nail?





Data:
m= 3kg, v= 10 m/s, d= 0.02 m, F=?
Principle
Initial = K.E Final = work
½ mv2=F·d
1
3kg  (10m / s ) 2  F  0.02m
2
2
1
m
3kg 100 2
2
s
F
 7500 N
0.02m
How fast will a 5 kg rocket be traveling if
its motor has a force of 2000 n and it
operates for 500 m?






Data
V=?, m=5kg, F= 2000 N, d=500m
Principle
Initial state = work Final state = K.E.
F·d = ½ mv2
V= 632 m/s
1
2
2000 N  500m   5kg  v
2
2  2000 N  500m
2
v
5kg
Potential Energy to Kinetic Energy


What is the speed of a ball that has fallen 2 m without
friction?
In the beginning all the energy is P.E.
–

After falling 2 m, all the energy is Kinetic
–




P.E. = mgh
K.E. = ½ mv2
K.E. = P.E.
½ mv2 = mgh
v = √2gh
v = √2·9.8 m/s2·2 m = √39.2 m2/s2 = 6.26 m/s
Density – help for your pre-lab



Density is the mass per unit volume.
Density = mass/volume
D=m/v
Mass
m
Density 

Volume V
Calculate the density of a wood cylinder of diameter 6
cm, length 10 cm and mass 250 grams.
Data: d = 6 cm, (r = 3 cm), L = 10 cm and m = 250 g.
volume  r L  3.14  (3cm) 10cm  283cm
2
2
mass
250 g
3
Density 

 0.885 g / cm
3
volume 283cm
3