Physics 101: Lecture 13
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Chapter 6 : Work and Energy
Quick Review of Last Time, Example Problems
Power, Work done by a variable force
Reminders:
Exam I, Tuesday, September 30th at 5 PM
See PHY101 Web page for room assignments
Please do not forget to bring your UB ID card !
Physics 101: Lecture 13, Pg 1
Work done by a constant Force
• W = F s = |F| |s| cos  = Fs s
|F| : magnitude of force
F

|s| = s : magnitude of displacement
s
Fs = magnitude of force in
direction of displacement :
Fs = |F| cos 
: angle between displacement and force
vectors
• Kinetic energy : KE= 1/2 m v2
• Work-Kinetic Energy Theorem:
KE = Wnet
Physics 101: Lecture 13, Pg 2
Work Done by Gravity
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Example 1: Drop ball
Wg = (mg)(S)cos
Yi = h0
S = h0-hf
Wg = mg(h0-hf) cos(00)
mg
S
y
= mg(h0-hf)
= PEinitial – PEfinal
Yf = hf
x
Physics 101: Lecture 13, Pg 3
Work Done by Gravity
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Example 2: Toss ball up
Wg = (mg)(S)cos
S = h0-hf
Wg = mg(h0-hf)cos(1800) =
=-mg(h0-hf)
= PEinitial – PEfinal
Yi = h0
mg
S
y
Yf = hf
x
Physics 101: Lecture 13, Pg 4
Work Done by Gravity
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Example 3: Slide block down incline
Wg = (mg)(S)cos
h0
Wg = mg(h/cos)cos

h
S = h/cos
hf
S
mg
Wg = mgh
with h= h0-hf
Work done by gravity is independent of path
taken between h0 and hf
=> The gravitational force is a conservative force.
Physics 101: Lecture 13, Pg 5
Concept Question
Imagine that you are comparing three different ways of having a ball
move down through the same height. In which case does the ball reach
the bottom with the highest speed?
1.
2.
3.
4.
Dropping
Slide on ramp (no friction)
Swinging down
All the same
correct
1
2
3
In all three experiments, the balls fall from the same
height and therefore the same amount of their
gravitational potential energy is converted to kinetic
energy. If their kinetic energies are all the same, and their
masses are the same, the balls must all have the same
speed at the end.
Physics 101: Lecture 13, Pg 6
Conservation of Mechanical Energy
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or
Total mechanical energy of an object remains constant
provided the net work done by non-conservative forces
is zero:
Etot = Ekin + Epot = constant
Ekin,f+Epot,f = Ekin,0+Epot,0
Otherwise, in the presence of net work done by
non-conservative forces (e.g. friction):
Wnc = Ekin,f – Ekin,0 + Epot,f-Epot,i
Physics 101: Lecture 13, Pg 7
Example Problem
Suppose the initial kinetic and potential energies of a system are 75J
and 250J respectively, and that the final kinetic and potential energies
of the same system are 300J and -25J respectively. How much work
was done on the system by non-conservative forces?
1. 0J
2. 50J
correct
3. -50J
4. 225J
5. -225J
Work done by non-conservative forces equals the
difference between final and initial kinetic energies
plus the difference between the final and initial
gravitational potential energies.
W = (300-75) + ((-25) - 250) = 225 - 275 = -50J.
Physics 101: Lecture 13, Pg 8
Power
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Average power is the average rate at which a net force
does work:
Pav = Wnet / t
SI unit: [P] = J/s = watt (W)
Or
Pav = Fnet s /t = Fnet vav
Physics 101: Lecture 13, Pg 9
Work done by a Variable Force
The magnitude of the force now depends on the
displacement: Fs(s)
Then the work done by this force is equal to the
area under the graph of Fs versus s, which can be
approximated as follows:
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W = S Wi = S Fs(si) s = (Fs(s1)+Fs(s2)+…) s
Physics 101: Lecture 13, Pg 10
Example Problems:
[C&J Chapter 6, problem 10] P = 256 N => Wnet = 0 and
Fnet,x = 0
 [C&J Chapter 6, problem 78]
Energy conservation (FN=0 => Wnc =0 ):
Etot,f= Etot,i

m g r cos  + ½ m vf2 = m g r (1)
Centripetal force is provided only by gravitational force
(FN=0) : - m g cos  = -m vf2/r (2)
=> Solve Eq.(2) for vf2 , plug in Eq.(1) and solve for :
cos  = 2/3 =>  = 48 degrees
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Physics 101: Lecture 13, Pg 11