Download Work and Energy

Work and
Energy
1
??Has Work Been Done??
1) A teacher applies a force to a wall and become exhausted
2) A book falls off a table and free falls to the ground
3) A waiter carries a tray full of meals above his head by one
arm across the room
4) A rocket accelerates through space
2
Work is defined as a force acting upon an object to cause a
displacement.
There are three key words in this definition
force, displacement, and cause.
In order for a force to qualify as having done work on an
object, there must be a displacement and the force must
cause the displacement.
3
If an object does not move then no work has been done.
Force and displacement must both be in the same direction for
work to have been done.
Example: A Flag in a parade goes horizontally but the
force being applied to the flag in order to hold it up is
vertical. Therefore no work is done.
4
??Has Work Been Done??
1) A teacher applies a force to a wall and
become exhausted
NO
2) A book falls off a table and free falls to
the ground
YES
3) A waiter carries a tray full of meals
above his head by one arm across the room
NO
4) A rocket accelerates through space
YES
5
Work Facts
The formula for work is simply force times distance.
W  fd
The unit for Work is Nm or J(Joule)
1 Joule is the work done by a Force of 1 Newton in
moving an object a distance of 1 meter.
6
Example: What Work is done by pushing a physics text book
with a Force of 20 N a Distance of 3m in an attempt to avoid
doing homework.
W  fd
W  203
W  60 Nm
7
Example:
Mr. Harper lifts up a 65kg student to a height of 0.50 m.
a) What Force does Mr. Harper use to lift the student?
First off: The applied force would be equal to the weight
therefore we need to find weight.
w  mg
w  659.8
w  637
f  640 N
8
b) What work does he do?
W  fd
W  6370.5
W  3185
.
W  320 J
9
Example: If two physics students are rearranging a room and they
decide to move a desk across the room, a total distance of 3.0 m.
If they move the desk at a constant velocity by each exerting
horizontal force of 200 N. Calculate the amount of work that was
done to move the desk across the room.
W  fd
W  4003
W  1200 J
W  12
. kJ
10
Example: A child ties a ball to the end of a 1 m long piece of
string and swings the ball in a full circle. If the string exerts
a continues force on the ball of 10 N, how much work does
the string due on the ball during one full revolution?
W  fd
W  100
W  0J
No distance. No work
11
Do
Practice Problems Pg 225 (pdf 34) #’s 4-10
12
Force vs Distance Graphs
If Mr. Harper was to push a square pig 4 m across the floor
with a constant force of 10 N. How much work would Mr.
Harper have done on the pig?
W  fd
W  104
W  40 J
13
Draw a force vs Distance graph of Mr. Harper pushing Peter the pig.
14
Calculate the area under the graph.
Area  Lenth  Width
A  410
A  40
What does this area represent?
The area under a force vs distance graph is equal to
the work done by the force.
15
Determine the amount of work done by
the changing force in the given graph.
2.5 J
2J
16 J
2+2.5+16+6=26.5 =27J
6J
16
Do
Practice Problems Pg 229 (pdf 34) #’s 11-12 omit 11(d)
17
Positive and Negative Work
Consider a weightlifter bench pressing a barbell weighing 650 N
to a height of 0.55 m. There are two distinct motions, the first is
when the barbell goes up and second is when the barbell is
lower back down. Calculate the work done by the weightlifter
during the two separate motions.
Work done going up
Work done going down
W  fd
W  fd
W   650  0.55 
W   650  0.55 
W  360 J
W  360 J
Total work done ( 0 J )
18
Do
Practice Problems Pg 235 (pdf 34) #’s 14-15
Section Review Pg 235 (pdf 34) #’s 1-5
19
Energy
There are many different forms of energy however all of
which are measured in units of Joules.
In this chapter we will look at two different forms of energy.
Kinetic energy and potential energy. As well as how they are
related to the concept of work.
http://www.youtube.com/watch?v=vl
4g7T5gw1M
20
Kinetic Energy
Kinetic energy: The energy of an object due to its motion.
Kinetic energy is directly proportional to the mass and the
velocity squared of a moving object.
1 2
Ek  mv
2
21
Example: A 0.200 kg hockey puck, initially at rest, is
accelerated to 27.0 m/s. Calculate the kinetic energy of the
puck both at rest, and in motion.
At Rest
1 2
Ek  mv
2
1
2
Ek   0.200  0 
2
Ek  0 J
In Motion
1 2
mv
2
1
2
Ek   0.200  27.0 
2
Ek  72.9 J
Ek 
22
Do
Practice Problems Pg 238 (pdf 35) #’s 19-21
23
The Work and Kinetic Energy Theorem
In order to do work on object there must be a force applied to the
object.
When a force is applied to an object it will accelerate.
When it accelerates there is an increase in velocity.
An increase in velocity will cause an increase in kinetic
Energy.
Therefore, the work done on object is equal to the change in
the kinetic Energy of the object.
W  Ek
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Example: A shot putter heaves a 7.26 kg shot with a final
speed of 7.51 m/s.
a) What was the kinetic
energy of the shot?
b) If the shot was initially at
rest how much work was done
on it to give its it this kinetic
energy?
1 2
Ek  mv
2
1
2
Ek   7.26  7.51
2
Ek  205 J
W  Ek
W  Ekf  Eki
W  205  0
W  205 J
http://www.youtube.com/watch?v=jmshPC
3zEPU
25
Example: A physics student does work on a 2.5 kg curling
stone by exerting a 40 N force to it horizontally over a
distance of 1.5 m.
a) Calculate the work done by the student on the stone.
W  fd
W   40 1.5 
W  60 Nm
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b) Assuming that the stone started from rest, calculate the
velocity of the stone at the point of release. Consider the ice
surface to be frictionless.
W  Ek
W  Ekf  Eki
W  1 mv 2f  0
2
2  60 
2W
vf 

m
2.5
v f  6.9 m
s
Example:
A 75 kg skateboarder initially moving at 8.0 m/s, exerts an
average force of 200 N by pushing on the ground, over a
distance of 5.0 m. Find the new kinetic energy of the
skateboarder.
W  Ek
fd  Ekf  Eki
fd  Eki  Ekf
Ekf   200  5   1
75  8 

2
2
Ekf  3.4kJ
28
Do
Practice Problems Pg 245 (pdf 35) #’s 22 - 26
Section Review Pg 246 (pdf 35) #’s 1 - 3
29
What do all of these
things have in common?
30
Potential Energy
Potential energy: is stored energy, or when an object has the
potential to do work.
There are many different types of potential energy such as a
battery, a waterfall, a compressed spring, gasoline or anything
that has the potential to do work.
In this chapter will concentrate on what is called gravitational
potential, or energy due to an objects position on earth.
Often we refer to what is called the total mechanical energy
of the system. Which is simply the total combined kinetic and
gravitational potential energies.
http://www.youtube.com/watch?v=9g6hUKx5xVc
31
Gravitational potential energy is directly proportional to an
object’s mass and height.
The higher an object is lifted the more gravitational potential
energy it will have.
Also a more massive object will have a larger gravitational
potential energy that a less massive object at the same height.
Gravitational potential energy can be found using the
following formula.
E g  mgh
32
Example: While setting up a tent you use a 3.0 kg rock to
drive the tent pegs into the ground. If you lift the rock to a
height of 0.68 m, what gravitational potential energy will the
rock have?
Eg  mgh
Eg   3.0  9.8 0.68
Eg  20 J
33
*** Caution ***
When talking about gravitational potential energy you have
to specify what the height is relative to.
ie: the ground, the table, the top of the hill, the bottom of
the hill, ect ....
34
Example:
A 2.0 kg textbook is lifted from the floor to a shelf 2.1 m
above the floor.
a) What is the gravitational
potential energy relative to the
floor?
b) What is the gravitational
potential energy relative to the
head of a 1.65 m tall person?
Eg  mgh
Eg   2  9.8 2.1
Eg  41J
Eg  mgh
Eg   2  9.8  2.1  1.65 
Eg   2  9.8  0.45 
Eg  8.8 J
35
Do
Practice Problems Pg 250 (pdf 36) #’s 27 & 29
36
Gravitational Potential Energy and Work
When you do work on an object by lifting it to a new
relative height. The object will as a result have an increase
in gravitational potential energy thus the work done on an
object is equal to the change in the gravitational potential
energy of the object.
W  Eg
37
Example:
A 65 kg rock climber did 16 kJ of work against gravity to
reach a ledge. How high did the rock climber ascend?
W  Eg
W  mgh f
W  Egf  Egi
W
hf 
mg
16000
hf 
 65 9.8
W  mgh f  mghi
hi  0
W  mgh f  0
W  mgh f
h f  25m
38
Question:
You carry a heavy box up a flight of stairs. Your friend
carries an identical box on an elevator to reach the same
floor as you. Which one, you or your friend, did the
greatest amount of work on the box against gravity?
Because the change in gravitational potential energy of
the two different boxes is the same, the work done on the
two boxes are equal.
39
So far we have discussed both the work/kinetic energy
theorem and the work/potential energy theorem.
W  Ek
W  Eg
In both cases the amount of work done on the system was
equal to the change in energy of the system.
As it turns out both theorems are a part of a single all
encompassing theorem called the work /energy theorem.
Where the work done on a system is equal the the change in the
total mechanical energy of the system.
W  E
40
Do
Practice Problems Pg 254 (pdf 36) #’s 30 -34
41
Power
42
Work has to do with a force causing a displacement.
Work has nothing to do with the amount of time that this force
acts to cause the displacement.
Power is the rate at which work is done. How fast is the work
being done.
It can be found using the following equation.
Work
Power 
Time
43
The metric unit of power is the Watt ( W ).
As is implied by the equation for power, a unit of power is
equivalent to a unit of work divided by a unit of time, thus, a
Watt is equivalent to a Joule/second
For historical reasons, the horsepower is occasionally used to
describe the power delivered by a machine. One horsepower is
equivalent to approximately 750 Watts.
44
Example:
In preparation for the next power outage Mr. Harper is
trying to set up a generator. If the generator is capable of
putting out 9.5 kW of power, what size of engine (in hp)
will he need in order to run the generator?
1000W 1hp
9.5kW 

 12.6666  13hp
1kW
750W
Example:
A crane is capable of doing 1.5 x 105 J of work in 10 s.
What is the power of the crane in watts?
W 1.5 10
P 
 1.5 104 W
t
10
5
46
Example: A cyclist and her mountain bike have a combined
mass of 60 kg. She’s able to cycle up a hill that changes her
altitude by 400 m in 1min.
a) How much work did she do against gravity in climbing
the hill?
W  fd  wh  mgh   60 9.8 400  240kJ
b) How much power is she able to generate?
W 2.4 105
P 
 3.9 103W
t
60
47
Example: Two physics students, Jacob and Ryan are in the weight
lifting room. It takes Jacob 3 sec to lift the 100 kg barbell over his
head a distance of 0 .75 m. It takes Ryan 2 sec to lift the same
barbell over his head a distance of 0.55 m.
a) Which student does the most work?
b) Which student delivers the most power?
48
Do
Practice Problems Page 266 (pdf 37) #’s 41 - 43
Horsepower Lab (pdf 37) Page 267
Read pg 268 (pdf 37) and make up your own
notes on efficiency
49
Efficiency
Efficiency is the ratio of useful energy or work output to the
total energy or work input.
Eo
Efficiency 
100%
Ei
Wo
Efficiency 
100%
Wi
50
Example: A model rocket engine contained explosives stores
3.50x103 J of chemical potential energy. When launched the
stored chemical energy is transformed into gravitational
potential energy. Calculate how efficiently the rocket
transforms the stored chemical energy into gravitational
potential energy if the 0.50 kg rocket is propelled to a height
of 100 m.
14%
51
Do
Practice Problems Page 270 (pdf 37) #’s 44 - 50
** EXTRA QUESTIONS **
(only if students want to)
End of Chapter Review Pg 274 (pdf 38)
#’s 1, 2, 4, 5, 6, 15, 16, 17,18, 20, 21, 23, 24, 25, 26, 27, 28
52
Conservation of Energy
It’s not just a good idea,
it’s the law!!
53
The Law of conservation of energy states that;
“Energy cannot be
created or destroyed
simply transformed.”
54
Example: A ball sitting on top of a table has gravitational
potential energy.
The ball then falls off the table and hits the ground.
Was all the gravitational potential energy lost?
Was all of the energy lost?
Explanation: As the ball falls from its initial position it
started to gain velocity.
This gain in velocity then results in a gain in kinetic energy.
So as the ball falls further, it goes faster, thus the
gravitational potential energy decreases and the kinetic
energy increases.
There has been a transfer of energy from gravitational potential
energy to kinetic energy. With no net energy loss.
56
57
Example: A square pig of mass 50 kg is released from rest
at a height of 10.5 m above the ground. With what velocity
would pig hit the ground?
Eg  mgh
Eg   50  9.8 10.5 
Eg  5145 J
Ek  1 mv
2
2 Ek
v
m
v  14 m
2
E g  Ek
v
2 Eg
m
2  5145 
v
50
s
Question: What caused the transformation of energy
(of the pig)?
Answer: The force of gravity
58
In the previous example what would happen if the pig
was to be pushed down a ramp so that his final velocity
would be zero when he reaches the bottom of the ramp?
Is energy conserved now?
At the bottom of the ramp the pig would have zero
gravitational potential energy and zero kinetic energy.
Therefore energy is not conserved.
Question: So where did the energy go?
Answer: All of the energy was lost due to friction.
59
For the above examples we have two different types of forces.
The first force, force of gravity, is what we call a conservative
force. The second force, the force friction is a non-conservative
force.
A conservative force (such as gravity) is a force that does not
remove energy from the system. It is also path independent.
This means that the path the object travels does not affect the
final energy of the system.
A non-conservative force (such as friction) is a force that
does remove energy from a system, and is path dependent.
The longer the path the more energy that will be removed.
60
The Law of Conservation of (Mechanical) Energy
When all the work done through a process is done by
conservative forces, the total mechanical energy of the system
after the process is equal to the total mechanical energy of the
system before the process.
The total energy before is equal to the total energy after.
Ei  E f
Eg  Ek  E  E
'
g
'
k
**The net energy is constant.**
**This however can not be said for non conservative forces.**
61
As the rock falls its potential energy is transformed by
gravity, a conservative force, into kinetic energy.
62
Question? Imagine an amusement park that has rides like
those illustrated below. The masses of the car and occupants
are the same for each case. Assume the wheels and track are
effectively frictionless. Each car starts from rest at level A.
What are the respective speeds of the cars when they reach
level B?
Answer: At level B all cars will have the same speed
63
Question? While playing catch by yourself, you throw ball
straight up. Neglecting friction how does the speed of the
ball, when returns to your hand, compared to the speed with
which it left your hand?
Answer: The ball will have the same speed except
going in the opposite direction.
64
Example: A skier is
gliding along with a
speed of 2.0 m/s at the
top of the ski hill, 40.0 m
high. The skier then
begins to slide down the
icy, frictionless, hill.
a) what will be the skiers
speed at a height of 25 m?
v '  17 m
Ei  E f
Eg  Ek  Eg'  Ek'
2
'
2 '
1
1
mgh 
mv  mgh 
m v
2
2  
2
'
2 '
1
1
gh 
v  gh   v 
2
2

v'  2 gh  gh'  1 v 2
2

2
v '  2  9.8 40    9.8  25   1  2  
2


s
v '  17.2626
65
b) what will be the height of the skier at a speed of 10.0 m/s?
Ei  E f
Eg  Ek  E  E
'
g
'
k
2
'
2 '
1
1
mgh 
mv  mgh 
m v
2
2  
2
2 '
1
1
gh  v   v 
'
2
2
h 
g
h' 
 9.8 40  1
2
1
2  10 
2 
2
9.8
h '  35.1020
h '  35m
2
Example: A person riding a bike approaches a hill with a speed
of 8.5 m/s. The total mass of the bike and rider is 85 kg.
a) The rider coasts up the Hill. Assume there is no friction, at
what height will the bike come to a stop?
Ei  E f
1 mv 2  mgh'
2
1 v 2  gh'
2
2
v
h' 
2g
Eg  Ek  E  E
'
g
'
k
0  Ek  Eg'  0
Ek  Eg'
h '  3.6862
h '  3.7 m
67
c) Does your answer depend on the mass of the bike and rider?
explain.
No, b/c mass cancels out
Example: Tarzan, mass of 85 kg, swings down from a tree
limb on the end of the 20.0 m vine. his feet touch the
ground 4.0 m below the limb.
a) How fast is Tarzan moving when he reaches the ground?
2 '
1
mgh 
m v
2  
2 '
1
gh 
v
2 
Ei  E f
Eg  Ek  Eg'  Ek'
Eg  0  0  E
v'  2 gh
Eg  E
v'  2  9.8 20 
'
k
'
k
v '  390 m
s
69
b) Does your answer depend on Tarzan’s mass? Explain.
No, because it cancels out.
c) Does your answer depend on the length vine? Explain.
No, because it is not needed in the calculation.
c) What is the only thing that the answer does depend on?
height
c) Does this make sense? Explain.
Yes, because the final velocity of a dropped object
only depends on the height from which it is dropped.
All masses fall with the same accelerations.
Example: A skier starts from rest at the top of the 45 m hill,
skis down a 30̊ incline into a valley continues up a 40 m Hill.
Both heights are measured from the Valley floor, assume you
can neglect to friction.
a) How fast is the skier moving at the bottom of the Valley?
2 '
1
mgh 
m v
2  
2 '
1
gh 
v
2 
Ei  E f
Eg  Ek  Eg'  Ek'
Eg  0  0  E
v'  2 gh
Eg  E
v'  2  9.8 45
'
k
'
k
v '  30 m
s
71
b) What is the skiers speed at the top of the next hill?
Ei  E f
Eg  0  E  E
'
g
'
k
2 '
1
mgh  mgh 
m v
2  
'
2 '
1
gh  gh   v 
2
'
v'  2  gh  gh' 
v'  2  9.8 45   9.8 40  
v '  9.9 m
s
Do
Practice Problems Page 287 (pdf 40)
#’s 1, 2, 3, 4, 6, 8
Section Review Page 300 (pdf 40)
#’s 1 - 6
Conservation of Energy Work Sheet
73