Download Work Energy and Power Workbook

Work, Energy and Power
Name:
Blk:
A Work Book for Chapter 6 of Physics by Giancoli 3rd ed.
This Workbook will only be marked based on completeness out of 100 marks.
Organization tidiness and the “look” of what you hand in is of utmost importance. It is
your responsibility to make sure that you understand the material. If you have any
questions be sure to ask me for help.
1) Read and understand section. 6-1 p.125 - 126.
2) Write down the first equation for work that you came to. In this equation, what do
the two little parallel lines mean?
3) Equation 6-1 has a cos  term.
a) What happens to the value of cosine as you move from 0 degrees to 90 degrees?
b) Explain what the purpose of the cosine term is.
4) Solve the following:
a) A 20 kg box is moved a distance of 20m to the right by a person exerting a force
of 120N . The coefficient of kinetic friction is 0.35
i) Sketch in all the forces acting on the 20 kg box.
ii) Calculate the work done by each of the forces.
iii) Calculate the net work done on the box.
iv) Calculate the net force
v) Calculate the acceleration.
120N
43 degrees
20 kg
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b) Do question “a” over again but this time incline the surface on which the box is
sliding up 10 degrees up from the horizontal.
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5) When you are doing questions involving the increase or decrease in the height of an
object, gravity provides the force. F = ma where a is g. i.e. F = mg.
a) So, W = Fd
W = mad
(you must be mad to do work) Where d is
the vertical distance that the object moves.
b) Read Example 6-2 and its solution p. 127. Explain how dcos = h
6) Explain how sin could be used to determine the change in the vertical distance.
7) Do questions 1 - 5 p. 144.
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8) Solve the following:
a) A 1200kg truck drives up a 3km long hill. a) If the truck uses 1.8 x 107 J of
energy to vertically climb 500m then how much energy does the truck waste? b)
If all the wasted energy is the result of friction, then what is the force of friction
acting against the movement of the truck? c) What is the efficiency of the truck?
Efficiency = (energy output/ energy input) x 100%
9) Do questions 6 - 10 p. 144 - 145.
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10) Read section 6.4 to the bottom of p. 131. You may recall from grade 9, 10, and 11 that
energy is neither created nor destroyed. This is the Law of Conservation of Energy.
It follows from this that if you do a certain amount of work to raise an object to a new
height then the work you do to get it there must be equal to the amount of energy that
it has gained - provided that the energy transfer is 100% efficient.
So, if you raise an object to a new height it gains energy. The energy that
objects have as a result of height above another position is their Gravitational
Potential Energy (GPE). Note that the gravitational potential energy depends
on the vertical height of an object above some reference level. This level is often
taken to be the ground, but this is not necessarily so. The equation for GPE is
therefore very similar to the work equation.
W = mad this equation becomes
GPE = mgh
Where m is mass, g is gravity, and h is the vertical distance between two
points. Note that the book uses y instead of h.
Click on the link below and try to answer the questions on the picture.
Kinetic vs. Potential Energy Diagram
Do example 6-6 on p. 132 Do it on your own before reading through the
solution.
11) Do questions 29 - 31 p. 145
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12) Read and understand section 6-3 p. 128 - 130
When you push an object you change its velocity therefore it must accelerate.
 v 22 - v12 
W = Fd = m ad = m 
 d
 2d 
Explain where the term in parentheses came from.
Simplifying the above equation you get
W  12 mv 22  12 mv 12
The term
1
2
mv 2 is defined as the Translational Kinetic Energy of an object.
So, KE =
1
2
mv 2
The only variable that can change in KE is v ; therefore, KE =
Hence, the change in kinetic energy KE =
1
2
m v 2 =
1
2
1
2
m( v 22  v12 )
m v 22  12 m v 12  W
13) Do questions 17 – 21, 23 - 27 odd p. 145
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14) Read and understand section 6-6 p.134.
15) Read and understand section 6-7 p. 135
Memorize that Law of Conservation of Energy
b) Explain the meaning of the following equation and give an example in which
you might use it to solve a problem.
1
2
mv12  mgy1  12 mv 22  mgy 2
16) Most conservation of energy problems can be solved using the following approach
i) Write down the following conservation of energy equation
PE + KE = PE’ + KE’ + OTHER
(OTHER is any energy that is lost or given off - heat and friction for example.)
ii) Decide which are not valid in the situation and if OTHER is required and
rewrite as needed. For example if there was no kinetic energy at the beginning
but there was heat loss due to friction you would rewrite as:
PE = PE’ + KE’ + Wfriction
Wfriction is the work done to overcome friction.
iii) Solve as required.
17) Do questions 33 and 37 p. 146
18) Read and understand sections 6-8 and 6-9 pp. 137 - 141.
i) Do questions 34, 35, 36, 38 - 42 p. 146
19) Read and understand section 6 -10 p. 141- 143.
i) Do questions 46 - 56 p. 146
20) Do questions 57, 60, and 67 p. 147 odd
21) Do every question that you got wrong the first time over again. I’m not kidding! This
will be your best learning tool. But don’t look at the answers before doing them.
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