AP Physics Chapter 8
Potential Energy
and Conservation
of Energy
1
AP Physics



2
Turn in Chapter 7 Homework, Worksheet, and
Lab
Lecture: Potential Energy and Conservation of
Energy
Q&A
Review on Work Done by Gravity
3

Work done by gravity (weight):
W  mgd cos   mg h

Interpretation:

Object falling down from rest  h < 0  Wg > 0

Wg = K  K > 0  K > 0 (Isn’t obvious?)

If you want to stop the object, K: +  0  K < 0
negative work
 W < 0 by you You are doing _______
 Object doing work to you
Object at height has the potential ability to do work (later)
Kinetic and Potential Energy

Kinetic energy: ability to do work due to motion
–
–

One object or system
Associated with an object
Potential energy: ability to do work due to a
relative position
–
System


–
4
Object and earth
Mass and spring
Associated with a force
Change in Potential Energy (U)
and Work

The change of a potential energy is defined to
equal to the negative of the work done by its
associated force.
U  W


5
Not directly defining U
W by gravity is the same for same h even
when object falls from different initial heights
Conservative and Nonconservative
Forces
Conservative Force:



The net work done by a conservative force on a particle moving
around any closed path is zero.
The work done by a conservative force on a particle moving
between two points does not depend on the path taken by the
particle.
Examples: gravity (weight), spring force
Nonconservative Force:


6
The work done by a nonconservative force on a particle moving
between two points depends on the path taken by the particle.
Example: friction, force you push a table from some initial point
to final point
Force and Potential Energy

Only conservative forces have potential
energies.



Nonconservative forces do not have potential
energies associated with them.


7
Gravitational potential energy
Spring potential energy
No pushing or pulling potential energy
No frictional potential energy
Gravitational Potential Energy,
U or Ug
y
U  Wg    mg h   mg h  mg  h f  hi   mgh f  mghi
Define Gravitational Potential Energy: U  mgy





m: mass
g = 9.8 m/s2
y: vertical position, height (upward has been defined
as the positive direction.)
Reference point: y = 0  U = 0
Free to choose reference point. (U has no physical
significance, only U has physical significance.)
Unit:
8
2
m
m
kg


m

kg

U

m
g
y

     
   N  m  J  K 
s2
s
Elastic or Spring Potential Energy,
U or Usp
1
1 2
1 2 1
2
2
kx

kxi

W


kx

kx

U 
f
sp
i
f 

2
2
2
 2

Define Spring Potential Energy:




9
1 2
U sp  kx
2
k: spring constant
x: displacement from equilibrium (relaxed)
position of spring
Reference point has been chosen to be at x = 0.
Unit:
N
2
U sp    k  x    m 2  N  m  J  W 
m
Example: Pg188-3
10
In Fig. 8-30, a 2.00 g ice flake is released from the edge of a
hemispherical bowl whose radius r is 22.0 cm. The flake-bowl
contact is frictionless.
a)
How much work is done on the flake by its
weight during the flake’s descent to the
bottom of the bowl?
Ice flake
b)
What is the change in the potential energy of
the flake-Earth system during that descent?
c)
If that potential energy is taken to be zero at
r
the bottom of the bowl, what is its value when
the flake is released?
d)
If, instead, the potential energy is taken to be
zero at the release point, what is its value
when the flake reaches the bottom of the
bowl?
Solution: Pg188-3
m  2.00  103 kg, r  0.220m
a ) h   r  0.220m, W  ?
W   mg h    2.00  103 kg  9.8m / s 2   0.220m   4.31 103 J
b) U  ?
U  W  4.31  103 J
c ) y  r  0.220m,U  ?
U  mgy   2.00  103 kg  9.8m / s 2   0.220m   4.31  103 J
d ) y   r  0.220m, U  ?
11
U  mgy   2.00  103 kg  9.8m / s 2   0.220m   4.31 103 J
Practice: Pg189-5
In Fig. 8-32, a frictionless roller coaster of mass m = 825 kg tops the
first hill with speed v0 = 17.0 m/s at height h = 42.0 m. How much
work does the gravitational force do on the car from that point to a)
point A, b) point B, and c) point C? If the gravitational potential
energy of the coaster-Earth system is taken to be zero at point C,
what is its value when the coaster is at d) point B and e) point A?
f) If mass m were doubled, would the change in the gravitational
potential energy of the system between points A and B increase,
decrease, or remain the same?
O
v0
A
B
h
h
h/2
12
C
O
v0
A
B
Solution:
Pg189-5
h
h
h/2
C
m  825kg , vo  17.0m / s, h  42.0m
a)O  A, h  0, W  ?
W   mg h  0
b)O  B, h 
h
h
 h   ,W  ?
2
2
 h  mgh

W   mg h   mg    
2
 2
m
 42.0m
2
s
 1.70  105 J
2
825kg  9.8
c)O  C , h  0  h   h, W  ?
13
m
W   mg h   mg   h   mgh  825kg  9.8 2  42.0m  3.40  105 J
s
Solution: Pg189-5 (continued)
h
d ) B : y  ,U  ?
2
h
mgy

mg
 1.70  105 J
U
2
e ) A : y  h ,U  ?
U  mgy  mgh  3.40  105 J
f )m  2m, U A B  ?
U AB  U B  U A 
mgh
mgh
 mgh  
2
2
m  2m  U AB  2U A B (Doubled)
14
Practice: Pg189-6
A 1.50 kg snowball is fired from a cliff 12.5 m high with an
initial velocity of 14.0 m/s, directed 41.0o above the
horizontal.
a) How much work is done on the snowball by he
gravitational force during its flight to the flat ground
below the cliff?
b) What is the change in the gravitational potential
energy of the snowball-Earth system during the flight?
c)
If that gravitational potential energy is take to be zero
at the height of the cliff, what is the value when the
snowball reaches the ground?
15
Solution Pg189-6 :
m
m  1.50kg , h  12.5m, vo  14.0 ,  o  41.0o
s
a)h  h  12.5m, W  ?
m

W  mg h  1.50kg  9.8 2   12.5m   184 J
s 

b) U  ?
U  W  184 J
c ) y  h  12.5m, U  ?
m

U  mgy  1.50kg  9.8 2   12.5m   184 J
s 

16
(Total) Mechanical Energy, E
Sum of kinetic and potential energy:
Emec  K  U
or simply,
E  K  U  K  Ug  Us
17
Conservation of Mechanical Energy
When only conservative forces are doing work within
a system, the kinetic energy and potential energy
can change. However, their sum, the mechanical
energy E of the system, remains unchanged.
 Ei  E f

 Ki  U i  K f  U f

 E  K  U  0

 K   U
18
Example: Pg199-116
A 70.0 kg man jumping from a widow lands in an elevated fire rescue net
11.0 m below the window. He momentarily stops when he has stretched
the net by 1.50 m. Assuming that mechanical energy is conserved
during this process and that the net functions like an ideal spring, find the
elastic potential energy of the net when it is stretched by 1.50 m.
Let y = 0 after the net is stretched 1.5 m, then
yi = 12.5 m, yf = 0, m = 70.0kg,
i
y = 1.5m + 11m
=12.5 m
Us.f = ?
Ei  E f
0
0
0
0
K i  U g .i  U s.i  K f  U g . f  U s. f
U g .i  U s. f
U s . f  U g .i
19
y = 1.50 m
f
y=0
m

 mgyi   70.0kg   9.8 2  12.5m   8575 J  8.58kJ
s 

Practice: Pg197-96
A volcanic ash flow is moving across horizontal ground when it
encounters a 10o upslope. The front of the flow then travels 920 m on
the upslope before stopping. Assume that the gases entrapped in the
flow lift the flow and thus make the frictional force from the ground
negligible; assume also that mechanical energy of the front of the flow
is conserved. What was the initial speed of the front of the flow?
Let y = 0 at bottom, then
yi = 0, yf = 920m sin10o = 160m, vf = 0,
f
vi = ?
Ei  E f
i
 Ki  U i  K f  U f
20
0 1
0
1
2
 mvi  mgyi  mv f 2  mgy f
2
2
1
 mvi 2  mgy f  vi 2  2 gy f
2
m
m
 vi  2 gy f  2  9.8 2  160m   56.
s 
s

y=0
k
Practice: Pg191-31
21
m

A block with mass m = 2.00 kg is placed against a spring on
a frictionless incline with angle  = 30.0o (Fig. 8-43).
(The block is not attached to the spring.) The spring,
with spring constant 19.6 N/cm, is compress 20.0 cm
and then released.
a) What is the elastic potential energy of the compressed
spring?
b) What is the change in the gravitational potential energy
of the block-Earth system as the block moves from the
release point to its highest point on the incline?
c)
How far along the incline is the highest point from the
release point?
L

i•
f •
h
y=0
Solution: Pg191-31
m  2.00kg , k  19.6
a )U s 
N  100 cm

cm  m

3 N

1.96

10
, x  0.200m

m

1 2 1
N
2
kx   1.96  103   0.200m   39.2 J
2
2
m
b)vi  0, v f  0, x f  0, U g  ?

0
0

K  U s  U g  0  U g   U s   U s. f  U s.i  U s.i  39.2J
c) L  ?
Wg   U g   mg h  mg h  U g
U g
39.2 J

 2.00m
 h 
2
mg
 2.00kg  9.8m / s

22
sin  

h
h
2.00m
 L

 4.00m
o
L
sin  sin 30.0
Practice: Pg190-18
A block of mass m = 2.0 kg is dropped
from a height h = 40 cm onto a spring of
spring constant k = 1960 N/m (Fig. 8-36).
Find the maximum distance the spring is
compressed.
m  2.0kg , h  0.40m, k  1960
xmax  ?
Let the maximum compression be x, and let y = 0 at
maximum compression, then
0
0 1
E f  Ei  U gf  U sf  U gi  U si  kx 2  mg  x  h   mgx  mgh
2
 kx 2  2mgx  2mgh  kx 2  2mgx  2mgh  0
x


23
2mg 
 2mg 
2
 4k  2mgh 
2k
mg 
 mg 
k
2

2mg 
 2mg 
2
 8kmgh
yi = x + h
i
h
y=x
f
2k
 2kmgh
2.0kg  9.8m / s 2  

 0.10m or 0.08m

2.0kg  9.8m / s 2   2 1960 N / m  2.0kg  9.8m / s 2  0.40m 
2
1960 N / m
N
,
m
k y=0
What if the block is slowly lowered to the top of
the spring? How much will it be compressed?
Conservation of Energy?
No, hand is doing negative work when lowering the block.
Then how?
F  0
i
h
N = Fsp
f
W
24
Force and Potential Energy
For each potential energy, there is a conservative
force associated with it.
dU
U
 F 
U  W   F x  F  
dx
x
Nonconservative forces have no potential
energies.
25
Work Done By External Force
Work done by external force will change the total
mechanical energy of the system:
Wext  Emec
If there is also kinetic friction,
Wext  W f  Emec  Wext  Emec  Eth
where Eth  W f  f k d is the heat gain (change in thermal energy)
Total Energy:
26
Etot  Emec  Eth  Eint
Eint = Chemical Energy & Nuclear Energy
Work and Energy Change
Wext  Emec
all external forces (excluding
spring force and gravity)
Total work done by __________________________.
Work-Kinetic Energy Theorem:
W  K
all external forces, normally including
spring force and gravity.
Total work done by _______________________________.
27
Practice: Pg198-105
The temperature of a plastic cube is monitored while the cube is
pushed 3.0 m across a floor at constant speed by a horizontal force of
15 N. The monitoring reveals that the thermal energy of the cube
increases by 20 J. What is the increase in the thermal energy of the
floor along which the cube slides?
d  3.0m, f k  Fext  15N , Eth.cube  20 J , Eth. floor  ?
Eth  W f     fk d   f k d 
Eth  Eth.cube  Eth. floor




f k d  Eth.cube  Eth. floor
 Eth. floor  f k d  Eth.cube  15N  3.0m  20 J  25J
28
Practice: Pg198-111
A 0.63 kg ball thrown directly upward with an initial speed
of 14 m/s reaches a maximum height of 8.1 m. What is the
change in the mechanical energy of the ball-Earth system
during the ascent of the ball to that maximum height?
m
m  0.63kg , vi  14 , v f  0, y  8.1m, Emec  ?
s
Emec  U  K  U   K f  Ki 
1
 mg h  mvi 2
2
2
m
1

 m
 0.63kg  9.8 2   8.1m    0.63kg   14   12 J
s 
2

 s
29
1
Practice:
Pg193-51
No friction
2
3
d
(k)
In Fig. 8-51, a 3.5 kg block is accelerated from rest by a
compressed spring of spring constant is 640 N/m. The
block leaves the spring at the spring’s relaxed length
and then travels over a horizontal floor with a
coefficient of kinetic friction k = 0.25. The frictional
forces stops the block in distance d = 7.8 m. What are
a) the increase in the thermal energy of the block-floor
system,
b) the maximum kinetic energy of the block, and
c)
the original compression distance of the spring.
30
1
2
Solution: Pg193-51
No friction
3
d
(k)
m  3.5kg , k  640 N / m, v1  0,   0.25, d  7.8m, v3  0
a )Eth  ?
m


0.25

3.5
kg
9.8
7.8m   67.J


Eth  W f  fd   Nd   mgd
2 
s 

b) K max  ?
During distance d (2 3),
During uncompressing of spring, (1 2)
 W  K  K 3  K 2   K 2
E1  E2  U1  K1  U 2  K 2
W f   K2
K2  W f  67 J
31
c) x1  ?
 Kmax  K2  67 J
1 2
U1  K 2  kx1  K 2
2
 x1 
2K2
2  67 J

 0.46m
k
640 N / m