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Schedule
Feb. 9
review
Feb. 11
review
Feb. 16
4.8-4.9
Feb. 18
5.1-5.2
Feb. 23
5.3,6.1-6.3
Feb. 25
6.4-6.7
March 1
6.8-6.10
March 3
review
Feb. 13
Exam 1
Feb. 20
boardwork
Quiz 4
Feb. 27
boardwork
Quiz 5
Exam 2: Friday, March 5.
March 5
Exam 2
A circular
motion
problem.
Example of Work: A block of mass M slides a distance D
along a frictionless inclined plane, which makes an angle 
with the horizontal. (a) What force(s) do work? (b) How much
work is done?
Litany for Work-Kinetic Energy Problems, step 1: draw
a basic representative sketch of the process.
D

M
Step 2: free-body diagram.
N
y
x
a
M
D

D
M
Mg

Forces: think gravity and contact.
I wanted to use W for weight, but
didn’t, to avoid confusion with Work.
Show displacement vector. (Better to not “touch” the mass.)
Show acceleration. Choose axes.
Show components of forces not parallel to axes. Better label
any angles!
Step 3: do not assume N = mg!
N
y
x
a
D
M
D

Mg
M

The Litany also says to label all non-perpendicular angles
between forces and displacements.
We have one such angle. We can’t call it . Let’s call it .
(Obnoxious special effect to help draw your attention to the
angle!)
Step 4: write down the appropriate OSE.
N
y
x
a
D
M
D

Mg
M

OSE:
[WF]if = FD cos .
Danger! The  in the OSE is the angle between the force and
displacement. An angle given as  in the problem may not be
the correct angle to use in the OSE!
That’s why I made you use different angle names in step 3.
The litany is designed to help you avoid painful mistakes.
After step 4, the Litany is no longer applicable.
The Litany is intended for use with problems involving work
and kinetic energy (which I have not yet introduced).
The Litany is still valid for setting up this problem.
Remind me again, what am I trying to calculate?
N
y
x
a
D
M
D

Mg
M

Oh, yes—which forces do work!
In this case N is perpendicular to the displacement. The angle
is 90˚ and cos(90˚)=0.
Special dispensation: any time a force F is perpendicular to the
displacement D, you may immediately write WF=0.
0
Thus, without further mathematics: WN.
N
y
x
a
D
M
D

Mg
M

The force due to gravity does work:
OSE:
[WF]if = FD cos .
[WMg]if = Mg D cos .
In the process of determining which forces did work, I also
answered part (b) of the problem…except that…
The angle  was not given in the original problem.
N
y
x
a
D
M
D

Mg
M

So I must write
[WMg]if = Mg D cos .
[WMg]if = Mg D cos (90-).
Now I am done! (And sin  would also be acceptable.)
Dang, there’s a lot of new stuff to learn here!
No, there’s not. Just
OSE:
[WF]if = FD cos 
which I give you anyway. The “hard” part is figuring out the
correct angle.
6.2 Work Done by a Varying Force
Sorry, I have to skip this section.
If you promise to read this section, I won’t test you on it.
In problems with varying forces, you really need to use
calculus. The ideas are the same as those you learn in this
class; the mathematics a bit more complex.
Naturally, I won’t give you a varying-force problem.
6.3 Kinetic Energy and the
Work-Energy Principle
What is kinetic energy?
Your text shows that the work to change the speed of a mass
m from v1 to v2 is
W = ½mv22 – ½mv12
so we define kinetic energy as
OSE:
K = ½mv2 .
With the above definition of K, we have
OSE:
[Wnet]if = K .
This is often called the work-energy theorem. It is one of the
BIG DEALS of physics.*
Note that (stuff) always means (stuff)final - (stuff)initial.
So K means Kfinal - Kinitial.
*Calling this a BIG DEAL will cause you to want to apply it to every problem in this
chapter. There are even BIGGER DEALS later that are usually better starting points
for problems.
Example 6-4. What is the KE of a baseball of mass m thrown
with a velocity of magnitude V? If the ball started from rest,
how much energy did it take to make it reach this speed?
Giancoli does the second part the “physicist way.” Let’s do it
the “official way” here (on the blackboard).
Let’s think for a minute about what I have done so far in
Chapter 6. First I introduced a definition of work.
Using this definition, we found
OSE:
[Wnet]if = K .
This is often called the Work-Energy Theorem.*
Now I will begin to introduce a REALLY BIG DEAL equation,
even bigger than the Work-Energy theorem. I’m going to
change the order of presentation a bit compared to Giancoli’s
development of the idea...
*Giancoli calls a slightly different form of the equation the “work-energy principle.”
6.5 Conservative and Nonconservative Forces
Conservative force: work done by the force does not depend
on path taken; only depends on initial and final positions (or
“states”).
Gravitational force is the prime example of a conservative
force.
Let me pick our text off the floor and put it on a table.
y
D
Fgrav
θ
yf
yi
Calculate the work done by gravity in moving the book from
the floor to the table (not precisely following the Litany):
1. Choose y-axis up, yi at floor, yf at table.
2. Indicate force due to gravity, displacement, and angle
between the two vectors (it’s the smaller of the angles).
OSE:
[WF]if = FD cos 
[Wgrav]if = Fgrav D cos 
3. But check out this trig:
y
D
yf-yi
Fgrav
θ
yf
yi
[Wgrav]if = Fgrav D cos 
[Wgrav]if = Fgrav D (-cos (180-))
[Wgrav]if = mg D ( - (yf-yi) / D )
[Wgrav]if = -mg(yf-yi)
Yes, gravity did negative work!
Let me digress on this negative work business…
[Wgrav]if = -mg(yf-yi)
In moving the book, I exert a force on the book, and in this
case, do positive work.
The force of gravity opposes me, and does negative work.
Your problems will often be set up so that friction also does
negative work.
Had I made the y-axis positive down, then yf<yi and gravity
does positive work (and I would have done negative work).
As we have seen before in this class, the sign of an answer
depends on the choice of axes.
[Wgrav]if = -mg(yf-yi)
The work done by gravity is independent of path; it depends
only on the starting and stopping locations.
“But wait a minute? What about the work you did in
lifting the book?”
Demonstrate: professor doing work on the book.
Clearly, the work I do on the book depends on the path I take.
However, the work gravity does on the book depends only on
the initial and final coordinates.
Gravity is a conservative force. The force exerted by my hand
is generally not a conservative force.
Friction is a prime example of a non-conservative force. Let’s
consider moving a book along a table. Looking down at the
tabletop.
B
D1
μN
A
D2
μN
μN
D3
C
Wfriction, ABC = - μND1 – μND2  μND3 = Wfriction, AC
Work by friction depends on path, so friction is a nonconservative force.
So why all this fuss about conservative and nonconservative
forces?
6.4 Potential Energy
KE is energy that an object possesses because it is in motion.
The work done on an object by a conservative force depends
only on the object’s initial and final position (or configuration).
It is thus possible to define a function (in the mathematical
sense) whose value is the work done on an object by a
conservative force. We call this function the potential energy.
Potential energy is “put into” or “taken out of” a system or
object by the action of a conservative force.
Nonconservative push
force sets swing in
motion.
high
Conservative
gravitational force
changes energy
between kinetic and
potential
mid
low
Potential energy is an object’s
energy due to its configuration
or surroundings.
In the case of gravitational force,
“configuration” refers to initial and final
positions.
The two conservative forces that we will
work with in this course are the force
due to a massless, frictionless spring,
and the force due to gravity.
These forces give rise to gravitational potential energy and
spring potential energy.
Potential Energy
We are going to use the letter U for potential energy.
For reasons which might become clear to you later, we define
OSE:
Uf – Ui = -[WC]if
Gravitational Potential Energy
Gravitational potential energy depends only on the vertical
distance between the two points of interest.
Recall that you can choose your y-axis and origin any where
you want. Gravitational potential energy differences are
independent of your choice of origin!
Recall that a bit earlier we found [Wgrav]if = -mg(yf-yi).
If we take yi=0 and yf=y, we have
OSE
Ugrav(y) = mgy
(near surface of earth, y-axis up)
Repeating the OSE, to make a point:
OSE
Ugrav(y) = mgy
(near surface of earth, y-axis up)
You always get to choose where y=0, and where Ugrav=0.
Choose it wisely to simplify your problem!
Spring Potential Energy
An unstretched spring “wants” to stay unstretched. A
stretched spring “wants” to go back to its equilibrium length.
According to Hooke’s law, the force that restores the spring to
its equilibrium length is proportional to the amount of stretch:
If s (x in the picture below) is defined as the difference
between the equilibrium (unstretched) length of the spring
and the stretched (or compressed length of the spring, then
the potential energy of the spring is
OSE
Uspring(s) = ½ks2
Your text derives this
equation on page 158.
Let’s take s>0 to mean the spring is stretched, and s<0 to
mean the spring is compressed (although because s is
squared, the sign doesn’t “matter” for our results).
Remember, the symbol V may be used to indicate potential
energy. Don’t get confused if I do that.
We’ll do examples with potential energies soon enough. For
now, let’s go back to section 6.5, which I didn’t quite finish.
6.5 Conservative and Nonconservative Forces
(continued)
Remember how our definition of potential energy,
Uf – Ui = -[WC]if
had that seemingly unnecessary minus sign in it?
Using our definition of U, we can do the following:
[Wnet]if = K
Wnet = Wconservative + Wnonconservative
Wconservative + Wnonconservative = K
-U + Wnonconservative = K
Wnonconservative = K + U.
Wnonconservative = K + U.
Giancoli calls the above equation the “work-energy principle.”
The next section (6.6) in Giancoli is a continuation of the
ideas developed in sections 6.4 and 6.5.
6.6 Mechanical Energy and its Conservation
We define the mechanical energy of a system to be
OSE
E=K+U.
With this definition, we can write the work-energy principle like
this:
OSE
Ef – Ei = [Wother]if .
If only conservative forces act on a system, the total
mechanical energy is conserved:
OSE
if [Wother]if = 0, then Ef – Ei.
This is the Law of Conservation of Mechanical Energy. It is a
REALLY BIG IDEA.
It’s a Hammer Equation!
If you have a great big nail to pound in, are you
going to pound it with a dinky little screwdriver?
Or a hammer?
Ef – Ei = [Wother]if
“What goes into Ef and Ei?”
K’s and U’s. Kinetic energies of any objects in our system.
Also potential energies.
“What kind of “things” do you know about that have potential
energies?”
Springs and Gravity!
Ef – Ei = [Wother]if
“OK, springs and gravity have potential energies. So what U’s
go into the E of the above OSE?”
Uspring and Ugrav! For every spring and mass in the system.
“Anything else?”
No, not until next semester.
Ef – Ei = [Wother]if
“If springs and gravity go into Ef and Ei, what goes into Wother?”
Work done by any force that doesn’t come from a spring or
gravity! DUH!
A look ahead: if we extend our definition of energy to include
forms other than mechanical, we find that the total energy of
a closed system is conserved. That’s a REALLY REALLY
BIG IDEA. You’ll have to wait a bit for it.
6.7 Problem Solving Using
Conservation of Mechanical Energy
I’ll work a simple example. A bowling ball of mass m is
dropped on a spring of force constant K from a height of H
above the spring. What is its speed after it has rebounded to
a height of H/2? Neglect friction and air resistance.
Anybody try to care to solve this using kinematics?
No, you don’t want to use kinematics. In fact, you can’t. You
don’t know how to handle the non-constant spring force.
But using energy methods makes this problem easy...