Transcript Physics 211 - University of Utah

Midterm 2 will be held on March 13. Covers units 4-9
Classical Mechanics
Lecture 9
Today's Concepts and Examples:
a) Energy and Friction
b) Potential energy & force
Unit 9 Homework Due
Thursday March 12 11:30 PM.
No extension!
Mechanics Lecture 9, Slide 1
Homework 8. Awesome Job!
Average = 92%
Mechanics Lecture 8, Slide 2
Practice Exams
Phys 1500 Exams https://utah.instructure.com/courses/320947/files
- Spring 2013: http://www.physics.utah.edu/~springer/phys1500/exams/MidtermExam2.pdf
- Solutions: http://www.physics.utah.edu/~springer/phys1500/exams/MidtermExam2Soln.pdf
- Long Sample: https://utah.instructure.com/courses/320947/files/45779670/download?wrap=1
Phys 2210 Exams
- Practice : http://www.physics.utah.edu/~woolf/2210_Jui/rev2.pdf
- Spring 2015: http://www.physics.utah.edu/~woolf/2210_Jui/ex2.pdf
Mechanics Lecture 8, Slide 3
Main Points
Mechanics Lecture 8, Slide 4
Main Points
Mechanics Lecture 8, Slide 5
Incline with Friction: Work-Kinetic Energy
Using Work-Kinetic Energy Theorem
Same result as using Newton’s Law
Mechanics Lecture 8, Slide 6
Work by Friction :Wfriction<0
What is the macroscopic work done on the
block by friction during this process?
A) mgH
B) –mgH
C) mk mgD
K  Wtot  Wgravity  Wfriction
0  Wgravity  Wfriction
0  mgH Wfriction
D) 0
must be negative
W friction  mgH
N1
W friction   m k mgD  m k mgD
m
N2
H
mg
m mg
mg
Mechanics Lecture 9, Slide 7
Checkpoint
What is the total macroscopic work done on the block by all forces
during this process?
B) –mgH
A) mgH
C) mk mgD
D) 0
K  Wtot
Ki  0
m
Kf 0
K  0  Wtot  0
H
D
Mechanics Lecture 9,
8, Slide 8
Mechanics Lecture 8, Slide 9
Force from Potential Energy:1D
Mechanics Lecture 8, Slide 10
Force from Potential Energy in 3-d
Gradient operator
Mechanics Lecture 8, Slide 11
Potential Energy vs. Force
dU ( x)
F ( x)  
dx
Mechanics Lecture 9, Slide 12
Potential Energy vs. Force
1 2
U ( x)   kx
2
dU ( x)
F ( x)  
dx
 kx
Mechanics Lecture 9, Slide 13
Potential Energy vs. Force
Mechanics Lecture 9, Slide 14
Potential Energy vs. Force
Mechanics Lecture 9, Slide 15
Potential Energy vs. Force
dU ( x)
F ( x)  
dx
Demo
Mechanics Lecture 9, Slide 16
Equilibrium
Mechanics Lecture 8, Slide 17
Equilibrium points
Mechanics Lecture 8, Slide 18
Equilibrium points
Mechanics Lecture 8, Slide 19
Equilibrium points
Mechanics Lecture 8, Slide 20
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 21
Block on Incline
W friction  
Wgravity
 
f k  dx   m k mg x cos 
 
  W  dx  mg x sin 
Mechanics Lecture 8, Slide 22
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  m k m g cosx2
x 2 
 T cosx1
T  m g sin   m k m g cos
Mechanics Lecture 8, Slide 23
Block on Incline
K  0  Wnet  ?
Mechanics Lecture 8, Slide 24
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 25
Gravitational Potential Energy
 
r  rE

rE

r
 
r  rM

rM
Mechanics Lecture 8, Slide 26
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 27
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 28
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 29
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 30
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 31
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 32
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 33
Trip to the moon
x
x
Mechanics Lecture 8, Slide 34
Block on Incline 2
Wtension
 
  T  dx  T cos x1
W friction  
 
f k  dx   m k 1mg x1
Mechanics Lecture 8, Slide 35
Block on Incline 2
Wnet  W friction  Wnormal  Wtension  W friction  Wtension  T cosx1  m k1m gx1
1
m(v 2f  vi2 )  Wnet  T cosx1  m k1m gx1
2
2(T cosx1  m k1m gx1 )
vf 
m
K 
Mechanics Lecture 8, Slide 36
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  m k m g cosx2
x 2 
 T cosx1
T  m g sin   m k m g cos
Mechanics Lecture 8, Slide 37
Block on Incline 2
Wgravity
 
  W  dx  mg sin x2
Mechanics Lecture 8, Slide 38
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)
U(x)
0%
0%
0%
0%
x
(a)
(b)
(c)
(d)
dU ( x)
F ( x)  
dx
Mechanics Lecture 8, Slide 39
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
0%
0%
0%
0%
U(x)
x
(a)
(b)
(c)
(d)
dU ( x)
F ( x)  
dx
Mechanics Lecture 8, Slide 40
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 biggest in the –x direction?
A) (a)
B) (b)
C) (c)
D) (d)
0%
0%
0%
0%
U(x)
x
(a)
(b)
(c)
(d)
dU ( x)
F ( x)  
dx
Mechanics Lecture 8, Slide 41