Transcript Work and Energy

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.
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Kinetic Energy
Kinetic energy: The energy of an object due to its motion.
Kinetic energy is directly proportional to the mass and the the
velocity squared of a moving object.
1 2
Ek  mv
2
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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 
3
Do
Practice Problems Pg 238 (pdf 34) #’s 19-21
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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
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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
2
1
W
mv  0
2 f
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
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Do
Practice Problems Pg 245 (pdf 35) #’s 22 - 26
Section Review Pg 246 (pdf 35) #’s 1 - 3
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What do all of these
things have in common?
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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.
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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.
Eg  mgh
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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
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*** 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 ....
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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
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Do
Practice Problems Pg 250 (pdf 36) #’s 27 & 29
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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
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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  E g
W  mgh f
W  E gf  E gi
W
hf 
mg
16000
hf 
 65 9.8
W  mgh f  mghi
hi  0
W  mgh f  0
W  mgh f
h f  25m
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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.
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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
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Do
Practice Problems Pg 254 (pdf 36) #’s 30 -34
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