Midterm November 8
Classical Mechanics
Lecture 8
Today's Concepts:
a) Potential Energy
b) Mechanical Energy
Mechanics Lecture 8, Slide 1
Average=
Average=89.6%
Mechanics Lecture 8, Slide 2
Lecture Thoughts
Mechanics Lecture 8, Slide 3
Integrals as Area Under a curve
http://hyperphysics.phy-astr.gsu.edu/hbase/integ.html
Mechanics Lecture 8, Slide 4
Potential & Mechanical Energy
Mechanics Lecture 8, Slide 5
Work is Path Independent for Conservative Forces
Mechanics Lecture 8, Slide 6
Work is Path Independent for Conservative Forces
Mechanics Lecture 8, Slide 7
Conservative Forces. Work is zero over closed path
Mechanics Lecture 8, Slide 8
Potential Energy
Mechanics Lecture 8, Slide 9
Gravitational Potential Energy
Mechanics Lecture 8, Slide 10
Mechanical Energy
Mechanics Lecture 8, Slide 11
Mechanical Energy
Mechanics Lecture 8, Slide 12
Conservation of Mechanical Energy
Mechanics Lecture 8, Slide 13
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 14
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 15
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 16
Finding the potential energy change:
Use formulas to find the magnitude
Check the sign by understanding the
problem…
Mechanics Lecture 8, Slide 17
Gravitational Potential Energy
Mechanics Lecture 8, Slide 18
Earth’s Escape Velocity
What is the escape velocity at the Schwarzchild radius of a black hole?
Mechanics Lecture 8, Slide 19
Clicker Question
A.
B.
C.
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Mechanics Lecture 8, Slide 20
Spring Potential Energy: Conserved
Mechanics Lecture 8, Slide 21
Vertical Springs
Massless spring
Mechanics Lecture 8, Slide 22
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 23
Non-conservative Forces
Work performed by non-conservative forces depend on exact path.
Mechanics Lecture 8, Slide 24
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 25
Summary
Mechanics Lecture 8, Slide 26
Spring Summary
M
kx2
x
Mechanics Lecture 8, Slide 27
Clicker, 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
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0%
2
1
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67% correct
3
h
Mechanics Lecture 8, Slide 28
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
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 29
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
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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 30
Clicker, Checkpoint
A.
B.
C.
D.
A block of mass m is launched up a frictionless ramp with an initial
speed v and reaches a maximum vertical height h. A second block
having twice the mass (2m) is launched up the same ramp with the
same initial speed (v). What is the maximum vertical height reached by
the second block?
A) h
B) 2 h
C) 2h
D) 4h
v
m
1 2
mgh  mv
2
h
1 2
h
v
2g
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Mechanics Lecture 8, Slide 31
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
55% correct
C) x2
 4x1
x
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0%
Mechanics Lecture 8, Slide 32
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 33
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 point o, which of the following quantities is at its maximum
value?
x
o
y
A) The block’s kinetic energy
B) The spring potential energy
C) Both A) and B)
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Mechanics Lecture 8, Slide 34
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 point x, which of the following quantities is at its maximum
value?
x
o
y
A) The block’s kinetic energy
B) The spring potential energy
C) Both A and B
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0%
0%
Mechanics Lecture 8, Slide 35
http://hyperphysics.phy-astr.gsu.edu/hbase/shm2.html#c4
Mechanics Lecture 8, Slide 36
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
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0%
Mechanics Lecture 8, Slide 37
http://hep.physics.indiana.edu/~rickv/SHO.html
A=0 at x=0!
Mechanics Lecture 8, Slide 38
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.
wrong
Compare K1 and K2.
A)
B)
C)
D)
K2 = 2K1
K2 = 4K1
K2 = 4K1/3
K2 = 3K1/2
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Mechanics
Mechanics Lecture
Lecture8,
8,Slide
Slide 39
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
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0%
Mechanics Lecture 8, Slide 40
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 41
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 42
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 GM e m
1
B) U case1  GM e m


 R 2R   2 2R
e 
e
 e
RE
Mechanics Lecture 8, Slide 43
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
 1
1  1 GM e m
 
 GM e m 
 Re 3Re  3 Re
B) U case 2
 1
1  2 GM e m
 
 GM e m 
 Re 3Re  3 Re
RE
Mechanics Lecture 8, Slide 44
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 45