Announcements
• Read Ch. 6 (Work & Energy): All of Chapter 6
• CAPA Set #8 due on Friday at 10 pm (shorter assignment)
• This week in Section, Assignment #5: Work & Energy
Print and bring to your Lab Section.
• Midterm Exam Tuesday October 11 at 7:30 pm
• Practice exam and solutions versions have a couple
differences. Apologies. Refer to solutions version.
• Office Hours… Nagle – Monday 2-3 pm Gamow F221
Uzdensky – Tuesday 11 am – 12 Help Room
Nagle/Kinney – Wed. 1:45-3:45 Help Room
Kinney – Thursday 11 am – 12 Help Room *
Exam #2
Tuesday, October 11
evening at 7:30-9:15 PM:
Reminders: Work and Kinetic Energy
Work – Force acting and resulting in displacement along
direction of force
Kinetic Energy – Energy of object in motion
1 2
KE  mv
2
Clicker Question
Room Frequency BA
Two marbles, one twice as massive as the other,
are dropped to the ground from the roof of a building.
(Assume no air resistance.)
Just before hitting the ground, the heavier marble has?
A)
B)
C)
D)
As much kinetic energy as the lighter one.
Twice as much kinetic energy as the lighter one.
Half as much kinetic energy as the lighter one.
Four times as much kinetic energy as the lighter
one.
E) Impossible to determine.
KE = (1/2) mv2, so if you double the mass
you double the KE.
Room Frequency BA
Clicker Question
A 1 kg mass is moved part way around a square loop as
shown. The square is 1 meter on a side and the final position
of the mass is 0.5 m below its original position. Assume that
g = 10 m/s2.
What is the work done by the force of gravity during this journey?
0J
A) 10 J
+10 J
-5 J
B) 5 J
C) 0 J
D) –10 J
E) –5 J
0J
Total work done W = (-5) + (0) + (+10) + (0) = +5 J
Work-Energy Principle
The work done by the net force on a single object
is equal to the
change in kinetic energy of that object:
Wnet = WFnet = ΔKE = KEf - KEi
Since work is a transfer of energy via a force, this is a
statement of conservation of energy.
Only applies if energy is not transferred to another form
(e.g. a change in potential energy, heat, chemical energy, etc.)
Work-Energy Principle
Wnet = WFnet = ΔKE = KEf - KEi
Push a book across a frictionless table with constant force Fext
#1: Fnet = Fext (since N and mg cancel)
#2: Fnet = ma  a = Fext / m
#3: W = +Fext Dx  Fext = W / Dx
#4: vf2 –vi2 = 2a Dx
vf2 –vi2 = 2(Fext/m) Dx
vf2 –vi2 = 2((W/Dx)/m) Dx
½ m(vf2 –vi2)= W
½ mvf2 –½mvi2 = DKE = W
Room Frequency BA
Clicker Question
A car of mass m is moving with speed v.
The driver applies the brakes and the car skids to a stop.
What was the magnitude of the work done by the
friction force on the tires?
|Wfric| = Ffric Δx = μK N Δx = |ΔKE|= ½ mv2
If the car’s speed were doubled,
the distance it would skid before stopping would be:
A) Doubled, B) Tripled, C) Quadrupled, D) Halved.
μK N Δx = ½
mv2
 m  2
Dx  
v

 2 N 
K
Potential Energy
PE is a stored energy associated with the
position or geometry of a physical system
Several varieties: gravitational, elastic, spring, ….
Definition:
PE is the amount of work done on a system by an
external force when Kinetic Energy does not change
and no heat energy flows in or out of the system.
ΔPE = Wext when ΔKE = 0
Again, effectively a re-statement of conservation of energy.
Demonstration
For a moment, they are at rest
near the top (KE = 0).
Later, they are moving quite
fast (large KE).
Where did the energy come
from since energy is
conserved?
Gravitational Potential Energy
Gravitational Potential Energy
Lift mass m
at a constant speed
by a height Δy
As height increases, PEgrav increases.
ΔPEgrav = Wext
ΔPEgrav = +Fext Δy
ΔPEgrav = +mg Δy
Often define PEgrav = 0 when y = 0.
PEgrav = mgy
Fnet = ma = Fext – mg=0
Fext = mg
PE is always defined relative to a
“reference level” where it is zero.
Mechanical Energy
Emechanical = KE + PE
Conservation of Mechanical Energy:
KE can change into PE and PE into KE, but the
total (KE+PE) is constant for an isolated system
with no lost energy (dissipation).
Emechanical = (KE + PE) = constant
DEmechanical = D(KE + PE) = 0
Investigate swing: http://phet.colorado.edu/sims/pendulum-lab/pendulum-lab_en.html
Conservation of Mechanical Energy
Emechanical = KE + PE = constant
(isolated system, no dissipation)
Consider mass m swinging attached to a string of length L.
The swing is released from rest at a height h.
What is the speed v of the swing when it reaches height h/2?
MEi  ME f
KEi  PEi  KE f  PE f
0
KE = ½ mv2
PEgrav = mgy
1 2
1 2
h
mv i  mgh  mv f  mg
2
2
2
1 2 1
v f  gh
2
2
v f  gh