Midterm next Friday November 8
Classical Mechanics
Lecture 9
Today's Concepts and Examples:
a) Energy and Friction
b) Potential energy & force
Mechanics Lecture 9, Slide 1
Main Points
Mechanics Lecture 8, Slide 2
Main Points
Mechanics Lecture 8, Slide 3
Mechanics Lecture 8, Slide 4
Force from Potential Energy:1D
Mechanics Lecture 8, Slide 5
Force from Potential Energy in 3-d
Gradient operator
Mechanics Lecture 8, Slide 6
Potential Energy vs. Force
dU ( x)
F ( x)  
dx
Mechanics Lecture 9, Slide 7
Potential Energy vs. Force
1 2
U ( x)   kx
2
dU ( x)
F ( x)  
dx
 kx
Mechanics Lecture 9, Slide 8
Potential Energy vs. Force
Mechanics Lecture 9, Slide 9
Potential Energy vs. Force
Mechanics Lecture 9, Slide 10
Clicker Question
A.
B.
C.
Suppose the potential energy of some object U as a function of x
looks like the plot shown below.
D.
Where is the force on the object zero?
A) (a)
B) (b)
C) (c)
D) (d)
60%
U(x)
30%
10%
0%
x
(a)
(b)
(c)
(d)
dU ( x)
F ( x)  
dx
Mechanics Lecture 8, Slide 11
Clicker Question
A.
B.
C.
Suppose the potential energy of some object U as a function of x
looks like the plot shown below.
D.
Where is the force on the object in the +x direction?
A) To the left of (b)
B) To the right of (b)
C) Nowhere
46%
38%
15%
0%
U(x)
x
(a)
(b)
(c)
(d)
dU ( x)
F ( x)  
dx
Mechanics Lecture 8, Slide 12
Clicker Question
A.
B.
C.
Suppose the potential energy of some object U as a function of x
looks like the plot shown below.
Where is the force on the object biggest in the –x direction?
A) (a)
B) (b)
C) (c)
D) (d)
17%
D.
67%
17%
0%
U(x)
x
(a)
(b)
(c)
(d)
dU ( x)
F ( x)  
dx
Mechanics Lecture 8, Slide 13
Equilibrium
Mechanics Lecture 8, Slide 14
Equilibrium points
Mechanics Lecture 8, Slide 15
Equilibrium points
Mechanics Lecture 8, Slide 16
Equilibrium points
Mechanics Lecture 8, Slide 17
Block on Incline
Wtension
 
  T  dx  T cos x1
Wnet  W friction  Wnormal  Wtension  Wtension  T cosx1
1
K  m(v 2f  vi2 )  Wnet  T cosx1
2
2T cosx1
vf 
m
Mechanics Lecture 8, Slide 18
Block on Incline
W friction  
Wgravity
 
f k  dx    k mg x cos 
 
  W  dx  mg x sin 
Mechanics Lecture 8, Slide 19
Block on Incline
Wtension
 
  T  d x  T x 2
Wnet  Wtension  Wgravity  W friction  K
2T cosx1
m
K  T cosx1
v f  0; vi 
 T cosx1  Tx2  m g sin x2   k m g cosx2
x 2 
 T cosx1
T  m g sin    k m g cos
Mechanics Lecture 8, Slide 20
Block on Incline
K  0  Wnet  ?
Mechanics Lecture 8, Slide 21
Energy Conservation Problems in general
For systems with only conservative forces acting
Emechanical  0
Emechanical is a constant
Emechanical  Ki  Ui  K f  U f  K (t )  U (t )
Mechanics Lecture 8, Slide 22
Gravitational Potential Energy
 
r  rE

rE

r
 
r  rM

rM
Mechanics Lecture 8, Slide 23
Gravitational Potential Problems
 
r  rE

rE
 
r  rM
 conservation of mechanical energy
can be used to “easily” solve
problems.
Emechanical  K  U 

r

rM
 Add potential energy from each
source.
GM E m
U Earth (rE )  
rE
U Moon (rM )  
1
mv (h) 2  U (h) gravity
2
 Define coordinates: where is
U=0?
U (r )  
GM E m
 0 as r  
r
GM M m
rM

GM E m GM M m
U total (r )       
r  rE
r  rM
Mechanics Lecture 8, Slide 24
Trip to the moon
1 2
m vi
2
GM E m
Ui  
RE
Ki 
Kf 0
Uf 
GM E m
Rf
Ki  U i  K f  U f
1 2 GM E m
GM E m
m vi 

2
RE
Rf
Rf 
 GM E m
 GM E
RE


1 2 GM E m 1 2 GM E
RE
m vi 
vi 
(1 
vi2 )
2
RE
2
RE
2GM E
Mechanics Lecture 8, Slide 25
Trip to the moon
U E ( RF )  
GM E m
RF
U M ( RF )  
GM M m
d ME  RF
MM
U M ( RE ) d ME  RF
M M RE


ME
U E ( RE )
M E (d ME  RF )
RE
RF  RE

Can ignore effect
of moon for this
problem at level
of precision for
SmartPhysics
U M ( RE )
 (0.01232)(0.01659)
U E ( RE )
 0.000204 0.02%
Mechanics Lecture 8, Slide 26
Trip to the moon
1 2 GM E m GM m m
GM E m GM m m
m vi 



2
RE
d ME
Rf
d ME  R f
 (d  x) b
b
c
c x   b(d  x)  cx 
  


 

x d  x  (d  x) x d  x  x   x(d  x) 
ax(d  x)  b(d  x)  cx
a
adx  ax2  bd  bx  cx  0
 ax  (ad  b  c) x  bd  0
2
a
1 2 GM E m GM m m
m vi 

2
RE
d ME
b  GM E m
c  GM M m
d  d ME
…or you can
practice solving
the quadratic
equation with
many terms!!!
Mechanics Lecture 8, Slide 27
Trip to the moon
d M E  d centers  RE
d E M  d centers  RM
1 2
m vi
2
GM E m GM m m
Ui  

RE
d M E
Ki 
1 2
m vf
2
GM E m GM m m
Uf 

d E M
RM
Kf 
Can NOT ignore
effect of moon for
this problem since
the rocket is AT
the moon in the
end !!!!
Mechanics Lecture 8, Slide 28
Trip to the moon
Ki  U i  K f  U f
1 2 GM E m GM m m 1 2 GM E m GM m m
m vi 

 m vf 

2
RE
d M E
2
d E M
RM
1
GM E GM m GM E GM m 

v f  2 vi2 



RE
d M  E d E M
RM 
2
 2GM E 2GM m
2GM E 2GM m 

v f  vi 1  2
 2
 2
 2
vi RE
vi d M  E vi d E M
vi RM 

 2GM E
M m RE
M m RE 
RE

v f  vi 1  2
(1 


) 
vi RE
M E d M  E d E M M E RM 

Mechanics Lecture 8, Slide 29
Trip to the moon
 2GM
M m RE
M R 
R
v f  vi 1  2 E (1 
 E  m E ) 
vi RE
M E d M  E d E  M M E RM 

M m RE
 0.000207
M E d M E
RE
d E M
 0.0167
M m RE
 0.0453
M E RM
2GM E
 1.020608
2
vi RE
Mechanics Lecture 8, Slide 30
Trip to the moon
Mechanics Lecture 8, Slide 31
Block on Incline 2
Wtension
 
  T  dx  T cos x1
W friction  
 
f k  dx    k 1mg x1
Mechanics Lecture 8, Slide 32
Block on Incline 2
Wnet  W friction  Wnormal  Wtension  W friction  Wtension  T cosx1   k1m gx1
1
m(v 2f  vi2 )  Wnet  T cosx1   k1m gx1
2
2(T cosx1   k1m gx1 )
vf 
m
K 
Mechanics Lecture 8, Slide 33
Block on Incline 2
Wtension
 
  T  d x  T x 2
Wnet  Wtension  Wgravity  W friction  K
2T cosx1
m
K  T cosx1
v f  0; vi 
 T cosx1  Tx2  m g sin x2   k m g cosx2
x 2 
 T cosx1
T  m g sin    k m g cos
Mechanics Lecture 8, Slide 34
Block on Incline 2
Wgravity
 
  W  dx  mg sin x2
Mechanics Lecture 8, Slide 35
Checkpoint
A.
B.
C.
D.
A block of mass m, initially held at rest on a frictionless ramp a vertical
distance H above the floor, slides down the ramp and onto a floor
where friction causes it to stop a distance r from the bottom of the
ramp. The coefficient of kinetic friction between the box and the floor
is k. What is the macroscopic work done on the block by friction
during this process?
40%
30%
B) –mgH
A) mgH
C) k mgD
D) 0
20%
10%
m
H
D
Mechanics Lecture 9,
8, Slide 36
Work by Friction :Wfriction<0
K  Wtot  Wgravity  Wfriction
Motionless box released on
frictionless incline which
then slides on horizontal
surface with friction to a
stop.  K  0
N1
0  Wgravity  Wfriction
0  mgH Wfriction
must be negative
m
N2
H
mg
 mg
mg
Mechanics Lecture 9, Slide 37
CheckPoint
What is the macroscopic work done on
the block by friction during this process?
B) –mgH
A) mgH
C) k mgD
D) 0
B) All of the potential energy goes to kinetic as it slides
down the ramp, then the friction does negative work to
slow the box to stop
C) Since the floor has friction, the work done by the block
by friction is the normal force times the coefficient of
kinetic friction times the distance.
m
H
D
Mechanics Lecture 9, Slide 38
Checkpoint
A.
B.
C.
D.
A block of mass m, initially held at rest on a frictionless ramp a vertical
distance H above the floor, slides down the ramp and onto a floor
where friction causes it to stop a distance D from the bottom of the
ramp. The coefficient of kinetic friction between the box and the floor
is k. What is the total macroscopic work done on the block by all forces
during this process?
B) –mgH
A) mgH
C) k mgD
D) 0
m
0%
0%
0%
0%
K  Wtot
H
D
Mechanics Lecture 9,
8, Slide 39
CheckPoint
What is the total macroscopic work done on
the block by all forces during this process?
B) –mgH
A) mgH
C) k mgD
D) 0
B) work= change in potential energy
C) The only work being done on the object is by the friction
force.
D) total work is equal to the change in kinetic energy. since
the box starts and ends at rest, the change in kinetic energy
is zero.
m
K  Wtot
H
D
Mechanics Lecture 9, Slide 40
unit 8
Average =97.9%
Mechanics Lecture 8, Slide 41
How much time did you spend on Homework 8
A.
B.
C.
D.
E.
0-30 minutes
31-60 minutes
61-90 minutes
91-120 minutes
More than 2 hours
Mechanics Lecture 8, Slide 42
Estimate your ability to do problems similar to
Unit 8 homework…in an exam setting
A.
B.
C.
D.
E.
F.
Excellent ability
Very good ability
Good ability
Fair ability
Poor ability
Not sure
Mechanics Lecture 8, Slide 43