Download The Potential Energy Function & Equilibrium

Elastic Potential Energy
Consider an ideal spring.

Fs  N 
Hooke’s Law


Fs  kx

x

x m 

Fs
Elastic Potential Energy Ee  - the negative of the area under the curve


of a graph of Fs vs. x
1 2
Ee  kx
2
Gravitational Potential Energy
A box of mass m is lifted from the ground
to a height y f at a constant velocity

by a constant applied force FA


Fg  N 
y

y m 

FA
Initial
Final
yyfi 00
m

Fg
FA  Fg

FA

?
Fg  constant,
negative
Gravitatio nal Potential Energy Eg  - the negative of the

Fg
area under the curve of the graph of


Fg vs. y (height)
Eg  mgy
The Potential Energy Function & Equilibrium

Potential Energy U  - The negative of the area under the curve of a graph of F  x 
U   Fdx
Consider the force of gravity:
For one dimensional,
conservative forces

Fg  mg
U g    mg  dy
U g  Eg  mgy
The Potential Energy Function & Equilibrium

Potential Energy U  - The negative of the area under the curve of a graph of F  x 
U   Fdx
For one dimensional,
conservative forces


Consider the spring: F  kx Hooke' s Law
s
U s    kx dx
1 2
U s  Ee  kx
2
U J 
xm 
The Potential Energy Function & Equilibrium
U J 
1 2
U s  Ee  kx
2
xm 
What does the slope of this graph represent?
dU
slope 
dx
dU d  1 2 
  kx 
dx dx  2

dU
 kx
dx
But


Fs  kx

dU
 F
dx

dU
F 
  slope of the U  x  graph
dx
The Potential Energy Function & Equilibrium
U J 
Notice that the slope is negative,
so the force is positive.
Notice that the slope is positive,
so the force is negative.
xm 

dU
F 
  slope of the U  x  graph
dx
Equilibrium - Occurs when F  0
Stable Equilibrium – occurs when a small displacement results
in a restoring force that accelerates the particle back
towards the equilibrium position.
The Potential Energy Function & Equilibrium
Consider the potential energy function for a skier on the top of hill between two valleys.
U J 
Notice that the slope is positive,
so the force is negative.
Notice that the slope is negative,
so the force is positive.
xm 

dU
F 
  slope of the U  x  graph
dx
Unstable Equilibrium – occurs when a small displacement
results in a force that accelerates the particle away
from the equilibrium position.
The Potential Energy Function & Equilibrium
Consider the potential energy function for an object on a plateau.
U J 
xm 

dU
F 
  slope of the U  x  graph
dx
Neutral Equilibrium – occurs when a small displacement results
in zero force so that the particle is again in
equilibrium.
The Potential Energy Function & Equilibrium
Example An object has the potential energy function as given below.
a. Sketch the corresponding Force vs. Position graph.
U J 
2
0
-2
-4
-6
-8
-10

F N 
3
2
1
0
-1
-2
-3

x m 
8 J
 2 N
4m
4J
1 N
4m
2
4
6
8
10
12
14

dU
F 
  slope of the U  x  graph
dx

x m 
2
4
6
8
10
12
14
The Potential Energy Function & Equilibrium
Example An object has the potential energy function as given below.
b. Identify all points of equilibrium.
U J 
2
0
-2
-4
-6
-8
-10

F N 
3
2
1
0
-1
-2
-3

x m 

x 5m
2
4
6
8
10
12
14

11 m  x  14 m Neutral Equilibrium

x m 
2
4
6
8
10
12
14
Stable Equilibrium
The Potential Energy Function & Equilibrium
Example An object has the potential energy function as given below.
Suppose the object has a total mechanical energy of  5 J as shown
by the double line on the U  x  graph.


c. What is the object' s kinetic energy at x  4 m and x  9 m?
U J 
2
0
-2
-4
-6
-8
-10
ET  U  Ek

x m 
Etotal
2
4
Total Mechanical
Energy

Ek  ET  U x 
6
8
10
12
14

From the graph of U  x 
U 4 m   8 J
U 9 m   6 J
Ek  5 J   8 J 
Ek  5 J   6 J 
Ek  3 J
Ek  1 J
The Potential Energy Function & Equilibrium
Example An object has the potential energy function as given below.
Suppose the object has a total mechanical energy of  5 J as shown
by the double line on the U  x  graph.

d. Between what values of x can the object move?
U J 
2
0
-2
-4
-6
-8
-10

x m 
Etotal
2

Ek  ET  U x 
1
and Ek  mv 2
2
4
6
8
10
12
14
Since the Ek is always positive the object

cannot reach positions where U  x   5 J .
Therefore, the object moves between

2.5 m  x  10 m