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Unit 5 Study Guide:
Definitions and Formulas:
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Energy is the property of an object that can produce a change in itself or its environment.
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Kinetic energy is the energy of an object that is changing its position (moving)
o KE = ½ mv2
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KE = Kinetic energy measured in Joules
m = mass measured in kg
v = velocity measured in m/s
Potential energy is energy that is stored in an object
o PE = mgh
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PE = potential energy measured in Joules
m = mass measured in kg
g = gravity measured in 9.8m/s2
h = height measured in m
Work is moving an object a distance by applying a force to that object
 If there is no distance, there is no work done.
 If dealing with lifting an object up, multiply the mass by gravity to get the Force needed
to use the W = Fd formula.
o W = Fd
o W = FcosΘd
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W = work measured in Joules
F = Force measured in N
d = distance measured in m
Power is the rate of doing work or the rate of using energy
o P=W
t
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P = power measured in Watts
W = work measured in Joules
t = time measured in seconds
Law of Conservation of Energy:
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Energy cannot be created nor destroyed. The total amount of energy in a system remains constant.
Energy can be transferred from one form to another.
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As the skier is at the top of his path, we calculate the potential energy to be 50,000 Joules. As the skier begins
to move down the mountain, the potential energy converts into kinetic energy. The energy always equals
50,000 J. At the bottom of the skier’s path, the kinetic energy is now equal to the potential energy at the top
of the path… 50,000 J. The total amount of energy throughout the entire path is a constant 50,000 Joules. This
is the law of conservation of energy.
Word Problems:
Example 1:
An object of mass 7.0 kg is dropped from a certain height. Just before it strikes the ground it has a
kinetic energy of 1750 J. From what height was the object dropped?
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Because of the law of conservation of energy, we can use the kinetic energy given in the word problem
and put it in the place of the potential energy to solve for height.
o PE = mgh
o 1750J = 7kg(9.8m/s2)h
o h = 25.5 m
Example 2:
How much work is required to lift a 100 kg crate uniformly through a vertical distance of 20 m?
 W = Fd
 First, we need to find the force.
o Fw = mg
o Fw = (100kg)(9.8m/s2)
o Fw = 980 N
 Now, plug in Force into W=Fd
o W = Fd
o W = (980N)(20m)
o W = 19,600 J
Example 3:
An object of mass = 2 kg is pulled by a constant force F = 4 N for a horizontal distance of 2 m. (Refer to Figure
above. What is the work done along the +x-axis?
 W = FcosΘd
 W = 4N(cos 30˚)(2m)
 W = 7J
Graphs:
Work = area. Separate the graph above into a triangle, a rectangle, and a triangle.
a. To find the work done from 0 to 5, use the first triangle in the graph.
- Area of a triangle: ½ base x height
½ (5)(3) = 7.5 J
b. To find the work done from 5 to 10, use the rectangle part of the graph.
- Area of a rectangle: base x height
(5) x (3) = 15 J
c. To find the work done from 10 to 15, use the second triangle in the graph.
- Area of a triangle: ½ base x height
½ (5)(3) = 7.5 J