Clicker Question
Room Frequency BA
A mass m is sliding on a frictionless plane tilted at an angle q
as shown. The mass is moving up the plane because it was
pushed up the plane by a spring a short time before (note
that the spring is no longer in contact).
Which of the following free-body diagrams most accurately
represents the forces on the mass m at the moment shown in
the diagram (after it was given an uphill push)?
59%
32%
1%
8%
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detailed information on “exam info” web link
Average = 67.4 out of 100
Stan. Dev. = 17.6
A range (85-100)
B range (70-85)
C range (55-70)
D range (40-55)
F range (0-40)
Clicker Question
Room Frequency BA
A mass m is pulled to the right along a frictionless surface by
a force of magnitude F at an angle q, as shown.
N
Fsinq
Fcosq
mg
Fnet,y=N+Fsinq-mg=0
N=mg-Fsinq
What is the magnitude of the normal force N exerted on
the mass by the table? 31%
3%
15%
3%
48%
This physics material is hard.
It is particularly difficult to overcome preconceived
notions that are backed up by years of observations.
“Do not try to bend the spoon — that's impossible.
Instead, only try to realize the truth: there is no spoon.”
No force is required in the direction of the velocity.
Keep working on these concepts and we are here to help.
Clicker Question
Room Frequency BA
Roller Coaster Problem
A
B
Assume an isolated system (no external work).
Assume no dissipation from friction (i.e. thermal energy generation)
Which of the following is true:
A) Speed at position A = Speed at position B
B) Speed at position A > Speed at position B
C) Speed at position A < Speed at position B
Clicker Question
Room Frequency BA
Roller Coaster Problem
N
Fnet,y = N-mg = may = 0
mg
If the car is just sitting at the top of this hill:
A) The net force on the car is upward.
B) The net force on the car is downward
C) The net force on the car is zero
Clicker Question
Room Frequency BA
Roller Coaster Problem
N Fnet,y = N-mg = may = mv2/r
v
mg
Approximately circular arc
If the car moving rightward at the top of this hill:
A) The net force on the car is upward.
B) The net force on the car is downward
C) The net force on the car is zero
General Statement of Conservation of Energy
ΔEmechanical + ΔEthermal = Wexternal
Emechanical = KE + PE
ΔEthermal = thermal energy generated = -Wfriction
Wfrict = -Ffriction Δx = -μk N Δx
Wexternal = external work done on the system
Wexternal
Can change energy
of the system
System
Energy can move from
KE to PE to Ethermal
General Statement of Conservation of Energy
ΔEmechanical + ΔEthermal = Wexternal
or
KEi + PEi + Wfrict + Wexternal = KEf + PEf
Wexternal > 0 if external work is done on the system
Wexternal < 0 if external work is done by the system
Wexternal = 0 if no external work is done on or by the system
Wfrict < 0 if there is friction
Wfrict = 0 if there is no friction
Clicker Question
Room Frequency BA
KEi + PEi + Wfrict + Wexternal = KEf + PEf
Under what condition will the final mechanical energy be
greater than the initial mechanical energy?
A)
B)
C)
D)
E)
Wexternal > |Wfriction|
Wexternal < |Wfriction|
Wexternal = |Wfriction|
It will never be greater than the initial mechanical energy
It will always be greater than the initial mechanical energy
Clicker Question
Room Frequency BA
KEi + PEi + Wfrict + Wexternal = KEf + PEf
Under what condition will Wexternal be negative?
A) If the external work is done on the system.
B) If the external work is done by the system.
C) It will never be negative.
D) It will always be negative.
If there is work done by the system,
the system loses energy to the
surroundings that it does work on.
One has to be careful about the definition of the “system”
System = everything in the box
Friction between hippo
and rug converts
mechanical energy to
thermal energy.
However, no external
work. All energy
remains within the
system (isolated).
System = everything in the box
Friction between hippo
and rug converts
mechanical energy to
thermal energy.
Friction force of rug on the
hippo is now an external
force. Thus energy leaves
the system.
Clicker Question
Room Frequency BA
A mass m slides down a rough ramp of height h. Its initial
speed is zero. Its final speed at the bottom of the ramp is v.
m
h
Which of the following expressions gives the speed squared (v2) of the
block when it reaches the bottom of the ramp?
A) 2gh
B) 2gh
C) 2gh  W frict
E) 2gh 
2W frict
m
D) 2gh 
KEi + PEi + Wfrict + Wexternal = KEf + PEf
2W frict
m
0 + mgh + Wfrict + 0
v  2 gh 
2
= ½ mv2 + 0
2W fric
m
Another Look
A mass m slides down a rough
ramp of height h. Its initial speed
is zero. Its final speed at the
bottom of the ramp is v.
h
Find the work done by friction…
KEi + PEi + Wfrict + Wexternal = KEf + PEf
0 + mgh + Wfrict +
0
= ½ mv2 + 0
Wfrict = ½ mv2 - mgh <0
D(Mechanical Energy) = -Wfric = mgh- ½mv2
Another Look
A mass m slides down a rough
ramp of height h. Its initial speed
is zero. Its final speed at the
bottom of the ramp is v.
q
h
Find the work done by friction using another method…
Wfric = Forcefric x displacement
Wfric = mk N x displacement
Wfric = mk (mgcosq) x (h/sinq)
Wfric = mk mgh cot(q) =DME = ½ mv2 - mgh
v  2 gh(1  m k cot(q )
The effect of some forces is expressed as a Potential Energy:
e.g., gravity, elastic forces produced by springs (later)
These forces are said to be Conservative.
They have the property that the work done depends only on
the starting and ending points, not the path taken while the
force is being applied.
The work done by some other forces depends on the path
taken: e.g., friction. Under the action of these forces,
mechanical energy is lost.
The friction force is Non-Conservative.
Its effect cannot be expressed as a Potential Energy.
Clicker Question
Room Frequency BA
A blue car coasts in neutral (no stepping on the gas or
brakes) down the blue path.
A red car coasts in neutral (no stepping on the gas or
brakes) down the red path.
If only gravity is acting, then
A) The red car ends up with a larger velocity
B) The blue car ends up with a larger velocity
C) The two velocities are equal at the end
Springs
Spring in “relaxed” position
exerts no force
Hooke’s “Law”
Force = -kx
Force
Compressed spring
pushes back
Force
Stretched spring
pulls back
k = “spring constant”
units = N/m
x = displacement of spring