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Energy
Potential
Energy
Gravitational
Potential
Elastic
Potential
Kinetic
Energy
Chemical
Potential
The basic motion of energy is stored work. A tankful of gas, a
heavy truck moving at speed, and a charged automobile battery
all possess energy.
§  The energy associated with a mass in motion is called kinetic
energy
§  Energy an object possesses due to its condition or position is
called potential energy.
Gravitational Potential Energy:
Energy an object possesses because of its position in a
gravitational field (height)
Formula
PE = m·g·h
m = mass (kg)
g = constant (9.81 m/s2 or 32 ft/s2)
h = vertical displacement height (m or ft)
Units: Joules (J) or Ft-lbs
Example 1:
A crate of mass 5,000 kg is raised slowly to a height of 12 m
above its original position. What is the potential energy at the
height of 12m?
PE = m·g·h
PE = 5000 kg x 9.81 m/s2 x 12 m
PE = 588,600 J
Elastic Potential Energy
Potential energy due to deformation (stretching) of an elastic
object.
1.  Drawn bow and arrow
2.  Stretched or compressed spring
3.  Stretched rubber band
Hooke's Law gives the relationship between the force applied
to an upstretched spring and the amount the spring is
stretched.
•  Every spring has an elastic limit.
•  If the spring is stretched within its elastic limit, it “springs”
back to its “rest point”.
•  If a spring is stretched beyond its elastic limit, it becomes
deformed.
•  As illustrated below, the distance a spring is
stretched is called the “elongation”.
•  Robert Hooke was the first to discover that the
spring force is directly proportional to the
elongation.
•  This relationship is known as Hooke’s Law.
Hooke’s Law formula:
F = k·x
F = the spring force (N)
k = the force or spring constant (N/m) ~ different for each spring
x = the distance or elongation of the spring from rest or equilibrium
position (m)
•  The force constant or spring constant is the force required to
stretch a given object.
•  The force or spring constant is different for different springs
and depends upon the type of material the spring is made of
as well as the thickness of the spring coil.
•  The greater the value of the force or spring constant, the
“stiffer” the spring.
Example #2
100 N/m
Example #3
(1.3, 5)
(1.5, 6)
(1.0, 4)
(0.67, 3)
(0.3, 1)
F = k·x
6 N = k ·1.5m
K = 4 N/m
Example #4
A 10 N force compresses a spring 0.25 meters from its
equilibrium position. Calculate the spring constant of this spring.
40 N/m
Example #5
A coiled spring is stretched 0.05 m by a weight of 0.50 N hung
from one end.
a.  How far will the spring stretch if a 1.0 N weight replaces the
0.50 N weight?
b.  What weight will stretch the spring a distance of 0.03 m?
1.  0.1 m
2.  0.3 N
Elastic PE Formula:
PE = ½ k (Δx)2
k = Force constant (N/m)
Δx = distance the object has stretched or been compressed from
rest position(m)
Units: Joules (J) or Ft-lbs
Example #6
A spring with a spring constant of 4 N/m is compressed by a
force of 1.2 N. What is the total elastic potential energy stored in
this compressed spring?
0.18 J
Example #7
Jan's mountain bike has a spring with a force constant of 64 N/m
in the front-wheel suspension. If it is compressed 0.17 m when
she hit a bump, how much energy does the front spring now
store?
PE = ½ k (∆x)2
PE = ½ (64 N/m)(0.17 m)2
PE = 0.925 J
Example #8
A spring has 1.1 J of potential energy and was compressed 0.2
m. What is its spring constant?
PE = ½ k (∆x)2
1.1 J = ½ k (0.2 m)2
k = 55 N/m
Chemical Potential Energy
Potential energy stored within the chemical
bonds of an object
•  Energy an object due to its motion
•  The ability or capacity of a moving object to move another
object
•  Kinetic energy exists whenever an object which has mass is
in motion with some velocity.
Kinetic Energy Formula:
KE = ½ mv2
m = mass (kg)
V = Velocity (m/s or ft/s)
Units: Joules (J) or Ft-lbs
Example #5:
What is the Kinetic Energy of a 1,000 kg car moving at 30 m/s?
KE = 450,000 J
Example #6:
What is the Kinetic Energy of a 20,000 kg freight car moving at 25
m/s?
KE = ½ mv2
KE = ½ (20,000) (25 m/s)2
KE = 6,250,000 J
Example #7:
What is the Kinetic Energy of a 0.04 g BB traveling 200 m/s?
KE = ½ mv2
KE = ½ (.00004 kg) (200 m/s)2
KE = 0.8 J
Example #8
Calculate the mass of a truck with 81,000 J of Kinetic Energy if its
speed is 14 m/s.
KE = ½ mv2
81,000 J = ½ m (14 m/s)2
m = 826.53 kg