Kinetic Energy: More Practice
Kinetic Energy: More Practice
Kinetic Energy: More Practice
Kinetic Energy: More Practice
Kinetic Energy: More Practice
Kinetic Energy: More Practice
Elastic Potential Energy:
Learning Goals
The student will describe the elastic potential energy of a
spring or similar object in qualitative and quantitative
terms and will investigate the transformation of
gravitational potential to elastic potential.
Elastic Potential Energy
4C Physics
Hooke’s Law
The stretch or compression of an elastic device
(e.g. a spring) is directly proportional to the
applied force:
Hooke’s Law
The stretch or compression of an elastic device
(e.g. a spring) is directly proportional to the
applied force:
stretch x 
Fx
or F x  kx
k
x  0 is the equilibriu m position
The spring constant
The constant k is called the spring constant or
force constant. It has units of N/m
and is the slope of the line in a force-extension
graph.
Example 1
A student stretches a spring 1.5 cm horizontally by
applying a force of magnitude 0.18 N.
Determine the force constant of the spring.
Example 1
A student stretches a spring 1.5 cm horizontally by
applying a force of magnitude 0.18 N.
Determine the force constant of the spring.
Givens :
x  0 . 015 m
F  0 . 18 N
Unknown :
k ?
Example 1
A student stretches a spring 1.5 cm horizontally by
applying a force of magnitude 0.18 N.
Determine the force constant of the spring.
Givens :
x  0 . 015 m
F  0 . 18 N
Unknown :
k ?
Select : F  kx  k 
F
x
Solve : k 
0 . 18 N
0 . 015 m
 12
N
m
Example 1
A student stretches a spring 1.5 cm horizontally by
applying a force of magnitude 0.18 N.
Determine the force constant of the spring.
Givens :
x  0 . 015 m
F  0 . 18 N
Unknown :
k ?
Select : F  kx  k 
F
x
Solve : k 
0 . 18 N
0 . 015 m
 12
N
m
Example 1
A student stretches a spring 1.5 cm horizontally by
applying a force of magnitude 0.18 N.
Determine the force constant of the spring.
Givens :
x  0 . 015 m
F  0 . 18 N
Unknown :
k ?
Select : F  kx  k 
F
x
Solve : k 
0 . 18 N
0 . 015 m
 12
N
m
Elastic Potential Energy
The force stretching or compressing a
spring is doing work on a spring,
increasing its elastic potential energy.
Note that this force is not constant but
increases linearly from 0 to kx. The
average force on the spring is ½kx.
Elastic Potential Energy
The force stretching or compressing a
spring is doing work on a spring,
increasing its elastic potential energy.
Note that this force is not constant but
increases linearly from 0 to kx. The
average force on the spring is ½kx.
W  F  d  Fx x 
1
2
2

1
2
kx  x 
1
2
kx
2
This kx is the elastic potential energy , E e .
Example 2
An apple of mass 0.10 kg is suspended from a vertical
spring with spring constant 9.6 N/m. How much elastic
potential energy is stored in the spring if the apple
stretches the spring 20.4 cm?
Givens :
k  9 .6
N
m
x  0 . 204 m
Unknown :
Ee  ?
Example 2
An apple of mass 0.10 kg is suspended from a vertical
spring with spring constant 9.6 N/m. How much elastic
potential energy is stored in the spring if the apple
stretches the spring 20.4 cm?
Givens :
Select : E e 
k  9 .6
Solve : E e 
N
m
x  0 . 204 m
Unknown :
Ee  ?
1
2
1
2
kx
2
9 . 6 0 . 204 m 
E e  0 . 20 J
N
m
2
Example 2
An apple of mass 0.10 kg is suspended from a vertical
spring with spring constant 9.6 N/m. How much elastic
potential energy is stored in the spring if the apple
stretches the spring 20.4 cm?
Givens :
Select : E e 
k  9 .6
Solve : E e 
N
m
x  0 . 204 m
Unknown :
Ee  ?
1
2
1
2
kx
2
9 . 6 0 . 204 m 
E e  0 . 20 J
N
m
2
Example 2 Follow-Up
How much gravitational potential energy did the
apple lose?
Example 2 Follow-Up
How much gravitational potential energy did the
apple lose?
Givens :
m  0 . 10 kg
g  9 .8
m
s
2
 h   0 . 204 m
Unknown :
E g  ?
Example 2 Follow-Up
How much gravitational potential energy did the
apple lose?
Givens :
Select :  E g  mg  h
m  0 . 10 kg
Solve :  E g  0 . 10 kg  9 . 8
g  9 .8
m
s
2
 h   0 . 204 m
Unknown :
E g  ?
 E g   0 . 20 J

m
s
2
 0 .204 m 
Example 2 Follow-Up
How much gravitational potential energy did the
apple lose?
Givens :
Select :  E g  mg  h
m  0 . 10 kg
Solve :  E g  0 . 10 kg  9 . 8
g  9 .8
m
s
2
 h   0 . 204 m
Unknown :
E g  ?
 E g   0 . 20 J

m
s
2
 0 .204 m 
The ideal spring
An ideal spring is one that obeys Hooke’s Law –
within compression/stretching limits. Beyond
those limits the spring may deform.
The ideal spring
An ideal spring is one that obeys Hooke’s Law –
within compression/stretching limits. Beyond
those limits the spring may deform.
Be gentle with
my springs!