Announcements
• Exam solutions on CULearn. Scores hopefully by
tomorrow and booklets returned next week.
• Read Ch. 6 (Work & Energy): Entire chapter.
• Read ahead Ch. 7 (Momentum): Ch. 7.1 – 7.6, 7.8.
• CAPA assignment #8 is due Friday at 10 pm
• This week in Section, Assignment #5: Work & Energy
Print and bring to your Lab Section.
• Standard Wednesday evening review session tonight
• Nagle help room hours today 1:45 – 3:45 pm
• Kinney help room hours Thursday 11 am - noon
1. Work:
Component of “specific” force along displacement x displacement
2. Kinetic Energy:
KE =
1 2
mv
2
3. Work – Energy Principle:
Wnet = WFnet = ΔKE = KEf - KEi
(No change in PE, no dissipation.)
4. Potential energy:
PE is the amount of work done on a system by an
external force when KE does not change and
no heat flows (no friction):
ΔPE = Wext when ΔKE = 0
Example: PEgrav = mgy
5. Conservation of “Mechanical Energy”:
Emechanical = KE + PE = constant
(isolated system, no dissipation)
Announcements
• Exam solutions on CULearn. Scores hopefully by
tomorrow and booklets returned next week.
• Read Ch. 6 (Work & Energy): Entire chapter.
• Read ahead Ch. 7 (Momentum): Ch. 7.1 – 7.6, 7.8.
• CAPA assignment #8 is due Friday at 10 pm
• This week in Section, Assignment #5: Work & Energy
Print and bring to your Lab Section.
• Standard Wednesday evening review session tonight
• Nagle help room hours today 1:45 – 3:45 pm
• Kinney help room hours Thursday 11 am - noon
1. Work:
Component of “specific” force along displacement x displacement
2. Kinetic Energy:
KE =
1 2
mv
2
3. Work – Energy Principle:
Wnet = WFnet = ΔKE = KEf - KEi
(No change in PE, no dissipation.)
4. Potential energy:
PE is the amount of work done on a system by an
external force when KE does not change and
no heat flows (no friction):
ΔPE = Wext when ΔKE = 0
Example: PEgrav = mgy
5. Conservation of “Mechanical Energy”:
Emechanical = KE + PE = constant
(isolated system, no dissipation)
Clicker Question
Room Frequency BA
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=0?
A) gh
KEi  PEi  KE f  PE f
B) 2 gh 0
C)
mv2
KE = ½
PEgrav = mgy
g
h
D) 2gh
1 2
+ mgh = mv  0
2
v  2gh
Clicker Question
Room Frequency BA
A projectile is fired with an initial speed vo at
an angle θ from the horizontal.
What is the KE of the projectile when it is on the way
down at a height h above the ground?
(Assume no air resistance)
Use PE=0 at ground level
A) ½ mvo2 + mgh
B) mgh
C) ½ mvo2 – mgh
D) Impossible to tell.
MEi  ME f
KEi  PEi  KE f  PE f
1
mv 02  0  KE f  mgh
2
1
KE f  mv 02  mgh
2
Clicker Question
Room Frequency BA
A mass slides down a frictionless ramp of height h.
Its initial speed is zero.
Its final speed at the bottom of the ramp is v.
h
As the mass descended, its KE
A) increased B) decreased C) remained constant
As the mass descended, its PE
A) increased B) decreased C) remained constant
As the mass descended, its (KE + PE) = total mechanical energy
A) increased B) decreased C) remained constant.
Clicker Question
Room Frequency BA
Mechanical Energy of a system is conserved
if there is no dissipation.
A physical system with no “dissipation”
is one in which:
Friction results in
energy transferred to
heat and thus loss of
mechanical energy.
(A) There is no friction.
(B) There is friction.
(C) No external forces are applied.
(D) External forces are applied.
(E) Energy is not conserved.
Clicker Question
Room Frequency BA
Mechanical Energy of a system is conserved
if the system is isolated.
An “isolated” physical system is one in which
External forces can do
(A) There is no friction.
work on the system.
(B) There is friction.
Thus energy is
(C) No external forces are applied. transferred into
(D) External forces are applied. or out of the system.
(E) Energy is not conserved.
Bowling Ball Demonstration
For the bowling ball to hit me, it would
be an example of violation of
conservation of energy.
PE = ME, KE = 0 (velocity = 0 at turning point)
PE = 0, ME = KE (maximum velocity)
Energy transformed into Kinetic Energy
Clicker Question
Room Frequency BA
A block initially at rest is allowed to slide down a frictionless
ramp and attains a speed v at the bottom.
How is the initial height h related to the
speed at the bottom of the ramp?
v2
A) h 
2g
B) h  2gv
C) h  g / v
v2
D) h 
g
PE = 0
Ei  E f
KEi  PEi  KE f  PE f
1 2
0  mgh  mv  0
2
v2
h
2g
Clicker Question
Room Frequency BA
A hockey puck slides without friction along a frozen lake toward an
ice ramp and plateau as shown. The speed of the puck is 4 m/s and
the height of the plateau is 1 meter.
+y
0
Will the puck make it all the way up the ramp?
A) Yes
B) No
C) Depends on the puck mass
ME(initial) = KEi + PEi = ½ mv2 + 0 = (8 x mass) Joules
ME(on the plateau) = KEp + PEp = ½ mvp2 + mgh
>= (9.8 x mass) Joules
Not enough energy to get up the ramp!