Transcript Physics 211 - University of Utah

Midterm 2 will be held on March 13. Covers units 4-9
Classical Mechanics
Lecture 8
Today’s Examples
a) Work and Kinetic Energy Problems
Today's Concepts:
a) Potential Energy
b) Mechanical Energy
Mechanics Lecture 8, Slide 1
Main Points
Mechanics Lecture 7, Slide 2
Main Points
Mechanics Lecture 7, Slide 3
Main Points
Mechanics Lecture 7, Slide 4
Work done by a Spring: Conservative Force!
Conservative Force
Net Work over closed path = 0
Physics 211 Lecture 7, Slide 5
Work done by Gravitational Force
Wgravity


  Fgravity  dl

GM E m
Fgravity  
rˆ
2
r
Mechanics Lecture 7, Slide 6
Work done by Gravitational Force
Wgravity


  Fgravity  dl

GM E m
Fgravity  
rˆ
2
r


GM E m
dW  Fgravity  dl  
rˆ  (drrˆ  rdˆ)
2
r
rˆ  (drrˆ  rdˆ)  drrˆ  rˆ  0  dr
Mechanics Lecture 7, Slide 7
Work done by Gravitational Force
dW  
GM E m
dr
2
r
r2
1
dr
2
r
r1
W12   dW  GM E m 
W12  GM E m(
1 1
 )
r2 r1
Mechanics Lecture 7, Slide 8
Work done by Gravitational Force
Mechanics Lecture 7, Slide 9
A.
Clicker Question
B.
C.
In Case 1 we send an object from the surface of the earth to a height
above the earth surface equal to one earth radius.
0%
0%
0%
In Case 2 we start the same object a height of one earth radius above
the surface of the earth and we send it infinitely far away.
In which case is the magnitude of the work done by the Earth’s
gravity on the object biggest?
A) Case 1
B) Case 2
C) They are the same
1 1
W  GM e m  
 r2 r1 
Mechanics Lecture 7, Slide 10
Clicker Question Solution
Case 1:
Case 2:
 1
1 
GM e m
W  GM e m
   
2R E
 2 RE RE 
1
1 
GM e m
  
W  GM e m 
2R E
  2 RE 
Same!
RE
2RE
Mechanics Lecture 7, Slide 11
Potential Energy
Mechanics Lecture 8, Slide 12
Checkpoint Clicker Question
A.
B.
C.
In Case 1 we release an object from a height above the surface of the earth equal toD. 1
earth radius, and we measure its kinetic energy just before it hits the earth to be K1.
In Case 2 we release an object from a height above the surface of the earth equal to 2
earth radii, and we measure its kinetic energy just before it hits the earth to be K2.
Compare K1 and K2.
wrong
A)
B)
C)
D)
K2 = 2K1
K2 = 4K1
K2 = 4K1/3
K2 = 3K1/2
0%
0%
0%
0%
Mechanics Lecture 8, Slide 13
A.
Clicker Question
For gravity:
B.
C.
GM e m
U (r )  
+U 0
r
What is the potential energy of an object
of mass m on the earths surface:
A) Usurface =
GMem
0
GMem
B) Usurface =
RE
GM
0%em
C) Usurface =
2RE
RE
0%
0%
Mechanics Lecture 8, Slide 14
A.
Clicker Question
GM e m
U (r )  
r
B.
C.
What is the potential energy of a object starting at the
height of Case 1?
GM e m
A) U1  
RE
B)
GM e m
U1  
2 RE
C)
GM e m
U1  
3RE
0%
RE
0%
0%
Mechanics Lecture 8, Slide 15
A.
Clicker Question
GM e m
U (r )  
r
B.
C.
What is the potential energy of a object starting at the
height of Case 2?
GM e m
A) U 2  
RE
GM e m
B) U 2   2 R
E
GM e m
C) U 2  
3RE
0%
RE
0%
0%
Mechanics Lecture 8, Slide 16
U surface
GM e m

RE
GM e m
U1  
2 RE
GM e m
U2  
3RE
What is the change in potential in Case 1?
 1
 1 GM e m
1
A) U case1  GM e m
 2 R  R   2 R
e 
e
 e
 1
1 
1 GM e m
B)U case1  GM e m


 R 2R    2 2R
e 
e
 e
RE
Mechanics Lecture 8, Slide 17
GM e m
GM e m
GM
m
e
U surface  
U2  
U1  
3RE
RE in potential2 RinE Case 2?
What is the change
What is the change in potential in Case 2?
A) U case 2
B)U case 2
RE
 1
1  2 GM e m
 GM e m
  
 3Re Re  3 Re
 1
1 
2 GM e m
  
 GM e m 
3 Re
 Re 3Re 
Mechanics Lecture 8, Slide 18
GM e m
U case1  
2 Re
What is the ratio
U case 2
2GM e m

3Re
K 2 U 2

K1 U1
2
4
3


1
3
2
A) 2
B) 4
C) 4/3
D) 3/2
Mechanics Lecture 8, Slide 19
Work on Two Blocks
Wgravity  (mgh)
Wgravity  m2 gd
x0  x
 
WN   N  dl
WN 
x0
x0  x
ˆj  iˆ  0  WN  0
 mgˆj  dxiˆ
x0
Mechanics Lecture 7, Slide 20
Work on Two Blocks
W  K
W  (m2 gh)
 (m2 gh) 
v1 f  v2 f 
WTension ,1  K1
K 
1
1
2
(m1  m2 )(v22 f  v20
)  (m1  m2 )v22 f
2
2
1
(m1  m2 )v22 f
2
 2m2 gh
(m1  m2 )
K1 
1
1
m1 (v12f  v102 )  (m1 )v12f
2
2
1
Wtension ,1  m1v12f
2
W1  Tx
W1
T
x
Mechanics Lecture 7, Slide 21
Work on Two Blocks

T2  Tˆj
x0  x
WTension , 2
 
  T  dl
x0
W2 

dl  dyˆj
 
T  dl  Tdy ( ˆj  ˆj )  Tdy  0
 WTension , 2  0
1
1
2
m2 (v22 f  v20
)  (m2 )v22 f
2
2
Mechanics Lecture 7, Slide 22
Work on Two Blocks 2
x0  x
 
W   F  dx  Fx
x0
x0  x
 
WN   N  dx  0
x0


1
1
m v 2f  v02  m v2f
2
2
2W
vf 
m1  m2
W  Fx 
Mechanics Lecture 7, Slide 23
Work on Two Blocks 2
W
1
m2 v 2f
2
W  Fx  T x
T
W
x
Wnet ,1  K1 
W1net 
1
m1v 2f
2
1
m1v 2f or W1net  Fnet x  ( F  T )x
2
Mechanics Lecture 7, Slide 24
Block Sliding
x0  x
Wspring
  x0  x   1
2
  F  dx   kx  dx  k x 
2
x0
x0
1 2
m vf
2
2Wspring
Wspring 
vf 
m
Mechanics Lecture 7, Slide 25
Block Sliding
x0  x
x  x

 0

W friction   F friction  dx     k mg iˆ  dx    k mg x 
x0
x0
W friction 
1
m(v 2f  v02 )
2
1

2
2   k mg x fric   k xspring  
2
 0
vf  
m
 k mg x friction   k xspring 2
1
2
1


2W friction  mv02 
2

vf  
m
x friction
1
2
k xspring 
2
 k mg
Mechanics Lecture 7, Slide 26
Block Sliding
x friction
1
2
k xspring 
2
 k mg
xspring 
Wspring 
2  k mg x friction 
k
1
2
k x 
2
Mechanics Lecture 7, Slide 27
Block Sliding 2
Wspring  K 
Wspring 
1
1
mv 2  mv 2f
2
2
1
1
2
k x   mv 2
2
2
x 
mv 2f
k
Mechanics Lecture 7, Slide 28
Block Sliding 2
Wspring 

1
1
mv 2  m v 2f  v02
2
2

W friction    k m gx
x 
W friction
 k m g
Mechanics Lecture 7, Slide 29
Block Sliding 2
W friction
 x patch 

   k mg 
2


Wspring  W friction
Wspring 
1
k (xnew ) 2
2
xnew 
 x patch 
1
   k (xnew ) 2
  k mg 
2
 2 
 x patch 

W friction    k' mg x patch     k mg 
2


 k' 
 k mgx patch
k
k
2
Mechanics Lecture 7, Slide 30
Block Sliding 2
Wfriction  k mgxrough
Mechanics Lecture 7, Slide 31
Potential & Mechanical Energy
Mechanics Lecture 8, Slide 32
Work is Path Independent for Conservative Forces
Mechanics Lecture 8, Slide 33
Work is Path Independent for Conservative Forces
Mechanics Lecture 8, Slide 34
Conservative Forces. Work is zero over closed path
Mechanics Lecture 8, Slide 35
Potential Energy
Mechanics Lecture 8, Slide 36
Gravitational Potential Energy
Mechanics Lecture 8, Slide 37
Mechanical Energy
Mechanics Lecture 8, Slide 38
Mechanical Energy
Mechanics Lecture 8, Slide 39
Conservation of Mechanical Energy
Mechanics Lecture 8, Slide 40
Relax. There is nothing new here
It’s just re-writing the work-KE theorem:
everything
except gravity
and springs
K  Wtot  Wgravity  Wsprings  WNC
 U gravity  Usprings
K  U gravity  Usprings  WNC
K  U  WNC
E  WNC
E  0
If other forces aren't doing work
Mechanics Lecture 8, Slide 41
Conservation of Mechanical Energy
Energy “battery” using
conservation of
mechanical energy
Upper reservoir
h
Pumped Storage Hydropower: Store energy by pumping water
into upper reservoir at times when demand for electric power is
low….Release water from upper reservoir to power turbines
when needed ...
http://www.statkraft.com/energy-sources/hydropower/pumped-storage-hydropower/
Mechanics Lecture 8, Slide 42
Gravitational Potential Energy
Mechanics Lecture 8, Slide 43
Earth’s Escape Velocity
What is the escape velocity at the Schwarzchild radius of a black hole?
Mechanics Lecture 8, Slide 44
Potential Energy Function
U0 : can adjust potential by an
arbitrary constant…that must
be maintained for all
calculations for the situation.
Mechanics Lecture 8, Slide 45
Finding the potential energy change:
Use formulas to find the magnitude
Check the sign by understanding the
problem…
Mechanics Lecture 8, Slide 46
Clicker Question
A.
B.
C.
W1  mgh   Mgh  K1
W2  mgh  2 Mgh  K 2  2K1
0%
0%
0%
Mechanics Lecture 8, Slide 47
Spring Potential Energy: Conserved
Mechanics Lecture 8, Slide 48
Vertical Springs
Massless spring
Mechanics Lecture 8, Slide 49
Vertical Spring in Gravitational Field
Formula for potential energy of vertical spring in gravitational field has same form
as long as displacement is measured w.r.t new equilibrium position!!!
Mechanics Lecture 8, Slide 50
Vertical Spring in Gravitational Field
Mechanics Lecture 8, Slide 51
Non-conservative Forces
Work performed by non-conservative forces depend on exact path.
Mechanics Lecture 8, Slide 52
Summary
K  Wtotal
U  W
E  K U
E  WNC
Lecture 7
Work – Kinetic Energy theorem
Lecture 8
For springs & gravity
(conservative forces)
Total Mechanical Energy
E = Kinetic + Potential
Work done by any force other than
gravity and springs will change E
Mechanics Lecture 8, Slide 53
Summary
Mechanics Lecture 8, Slide 54
Spring Summary
M
kx2
x
Mechanics Lecture 8, Slide 55
Checkpoint
A.
B.
C.
D.
Three balls of equal mass are fired simultaneously with
equal speeds from the same height h above the ground.
Ball 1 is fired straight up, ball 2 is fired straight down, and
ball 3 is fired horizontally. Rank in order from largest to
smallest their speeds v1, v2, and v3 just before each ball
hits the ground.
A) v1 > v2 > v3
B) v3 > v2 > v1
C) v2 > v3 > v1
D) v1 = v2 = v3
87% correct
2
1
3
h
Mechanics Lecture 8, Slide 56
CheckPoint
A) v1 > v2 > v3
B) v3 > v2 > v1
C) v2 > v3 > v1
D) v1 = v2 = v3
2
1
3
h
E  K  U  0
87% correct
They begin with the same height, so they have the same potential energy and
change therein, resulting in the same change in kinetic energy and therefore the
same speed when they hit the ground.
The total mechanical energy of all 3 masses are equal. When they reach the
same height their kinetic energies must be equal, hence equal velocities.
Mechanics Lecture 8, Slide 57
Clicker Question
A.
B.
C.
Which of the following quantities are NOT the same
for the three balls as they move from height h to
the floor:
2
1
3
h
0%
0%
0%
A) The change in their kinetic energies
B) The change in their potential energies
C) The time taken to hit the ground
Mechanics Lecture 8, Slide 58
A.
Clicker Checkpoint
B.
C.
A box sliding on a horizontal frictionless surface runs into a fixed spring,
compressing it a distance x1 from its relaxed position while momentarily
coming to rest.
If the initial speed of the box were doubled, how far x2 would the spring
compress?
A)
x2  2x1
B) x2
 2x1
90% correct
C) x2
 4x1
x
0%
0%
0%
Mechanics Lecture 8, Slide 59
CheckPoint
x
1 2
KE  mv
2
1 2
PE  kx
2
A)
x2  2x1
m v1
x1  x(v  v1 ) 
k
B) x2
 2x1
C) x2
 4x1
2
m(2v1 ) 2
x2  x(v  2v1 ) 
k
x2

x1
m(2v1 ) 2
( 2) 2
k

2
2
1
m(v1 )
k
Mechanics Lecture 8, Slide 60
http://hyperphysics.phy-astr.gsu.edu/hbase/shm2.html#c4
Mechanics Lecture 8, Slide 61
Clicker Question
A.
B.
C.
A block attached to a spring is oscillating between point x (fully
compressed) and point y (fully stretched). The spring is un-stretched at
point o. At which point is the acceleration of the block zero?
x
o
y
A) At x
B) At o
C) At y
0%
0%
0%
Mechanics Lecture 8, Slide 62
http://hep.physics.indiana.edu/~rickv/SHO.html
A=0 at x=0!
Mechanics Lecture 8, Slide 63
Integrals as Area Under a curve
http://hyperphysics.phy-astr.gsu.edu/hbase/integ.html
Mechanics Lecture 8, Slide 64
Homework
Average=85%
Average=86%
Average=82% including 2 who
did not do the homework
Mechanics Lecture 8, Slide 65
Example Test Problem
1. What is the work done by gravity during its slide to the bottom of the ramp?
2. What is the work done by friction during one pass through the rough spot?
3. What is the maximum distance the spring is compressed by the box?
4. What is the maximum height to which the box returns on the ramp?
Mechanics Lecture 6, Slide 66
Lecture Thoughts
Mechanics Lecture 8, Slide 67