Download Total mechanical energy

Energy Unit Study Guide
Energy Types
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What is energy? In physics, energy is the ability to do work, or use a force to move
an object some distance.
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Types of energy
1. Kinetic Energy (KE)
2. Potential Energy (PE)
a. Gravitational Potential Energy (GPE)
b. Elastic Potential Energy (EPE) or Spring Potential Energy (SPE)
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Kinetic energy (KE)
• The energy of motion.
• Depends on the speed of the object and its size.
• Equation:
KE = 1/2 mv2
where m = mass, v = velocity
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Gravitational Potential Energy (GPE)
• Energy of position.
• Depends on the size of the object and height off the ground.
• Equation:
GPE= mgh
where m = mass (kg), g = acceleration due to gravity (g = 9.8 m/s2 on
Earth), and h = height (m)
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Example: A 75 kilogram rock is on top of a 4 meter high post. What is the
potential energy of the rock on top of the post?
Wanted:
GPE
Given:
1. m = 75kg; 2. h = 4m; g = 9.8m/s2
Equation:
GPE= mgh
Substitution:
GPE=(75)(9.8)(4)
Math/Answer:
GPE=2940 J
Elastic Potential Energy or Spring Potential Energy (EPE/SPE)
• Depends on the deformation of the object (stretching, compressing or
bending).
• Equation:
SPE= 1/2 kx2, where k=spring constant, x=distance
spring is stretched/compressed (m)
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Summary: What type of energy does an object have?
 Is it moving?
o If yes, it DOES have kinetic energy.
 Is the object above the ground?
o If yes, it does have gravitational potential energy.
 Is there a spring compressed or stretched?
o If yes, it does have spring potential energy.
Total mechanical energy (TME)
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Total mechanical energy (TME or ME) is the sum of all of the mechanical energy in
an object or set of objects (system).
TME=KE+GPE+SPE
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Total mechanical energy is the energy possessed by an object due to either its
motion or its stored energy of position.
The total amount of mechanical energy is merely the sum of these three forms of
energy.
An object with mechanical energy is able to do work on another object.
Conservation of Energy
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Definition of a system: A system is a portion of the universe we have chosen to
study.
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Three (3) types of systems.
Open system: An open system can exchange matter
and energy through its boundary.
Closed system: We call it a closed system if no matter
is coming in or out.
The net external force must be zero
for this to happen.
Isolated system: In an isolated system, neither matter
nor energy cross the boundary. The
system is completely free of outside
influence.
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Examples of systems
• A sealed thermos (heat cannot go in or out)
• The Earth (ignore meteorites and astronauts)
• This room after the bell rings
 Isolated
 Closed
 Open
Conservation of energy: In physics, the law of conservation of energy states that:
1. the total energy of an isolated system cannot change - it is said to be conserved
over time.
2. Energy can be neither created nor destroyed,
3. but can change form, for be transformed, from one form to another (GPE to KE
for example) instance chemical energy can be converted to kinetic energy in the
explosion of a stick of dynamite.
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The math of conservation of energy (isolated systems only!)
TME = KE + GPE + SPE
TME initial (TMEi) = TME final (TMEf)
TMEi = TMEf
KE i + GPE i + SPE i = KE f + GPE f + SPE f
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TME Example Problems:
1. During a flood a tree trunk of mass 100 kg falls down a waterfall. The
waterfall is 5 m high. If air resistance is ignored, draw and calculate:
a. Draw a picture for this situation: include the waterfall, height, mass.
b. the potential energy of the tree trunk at the top of the waterfall.
c. the kinetic energy of the tree trunk at the bottom of the waterfall.
d. the magnitude of the velocity of the tree trunk at the bottom of the
waterfall.
2. A tennis ball, of mass 0.120 kg, is dropped from a height of 5 m. Ignore air
friction.
a. Draw a picture of this situation
b. What is the potential energy of the ball when it has fallen 3m ?
c. What is the velocity of the ball when it hits the ground?
ME - Example Problems: Use the Please Show Your Work (PSYW) Method!
Equations: g=9.8 m/s2 ; KE=½ mv2 ; GPE=mgh ; TME=KE+GPE+SPE
1. What is the kinetic energy (KE) of a 12kg mass moving at 7 m/s?
Wanted:
Given:
KE
1. m = 12kg
2. v = 7 m/s
Equation(s): KE=½ mv2
Sub.:
KE = ½ (12) (7) 2 (note: units not shown to keep math easier
Math:
= ½ ( 12)(49)
to read.)
= 249
Answer:
KE = 249 J
2. What is the gravitational potential energy (GPE) of a 50kg block that is 40
meters off the ground?
Wanted:
Given:
GPE
1. m = 50kg
2. h = 40m
Equation(s): GPE=mgh
Sub.:
GPE = (50)(9.8)(40)
Math:
= 19,600
Answer:
GPE = 19,600 J
3. What has more KE, an 9kg ball moving at 12m/s or a 3kg rock moving at 21
m/s?
Wanted:
Given:
What has more KE, the ball or the rock
1. mB = 9kg
2. vB = 12 m/s
3. mR = 3kg
4. vR = 21m/s
Equation(s): 1. KE=½ mv2 (Use for both the rock and the ball)
2. Then compare the two values and select the larger KE
Sub.:
Ball: KEB = (0.5)(9)(12)2 Rock: KER = (0.5)(3)(21)2
Math:
= (0.5)(9)(144)
= (0.5)(3)(441)
= 648 J
= 661.5 J
Answer:
The rock with 661.5 J has more KE
4. How high do you have to lift a 2.5kg tool box to give it 73.5 Joules (J) of
energy?
Wanted:
Given:
h
1. m = 2.5kg
2. GPE = 73.5 J
Equation(s): GPE=mgh
Sub.:
73.5 J = (2.5kg) (9.8 m/s2) (h)
Math:
____73.5____ = h
(2.5)(9.8)
73.5 / 24.5 = 3 = h
Answer:
h = 3m
5. An object with a mass of 15 kg has 300J of kinetic energy. What is its
velocity?
Wanted:
Given:
v
1. m = 15kg
2. KE = 300 J
Equation(s):
KE = ½ mv2
Sub.:
300 = (0.5) (15) v2
Math:
300 = 7.5 v2
300 / 7.5 = v2
40 = v2
√40 = √v2
Answer:
6.3 m/s = v
Work
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Definition of Work: When a force acts upon an object to cause a displacement of the
object, it is said that work was done upon the object.
There are three key ingredients to work:
1. force,
2. displacement, and
3. cause.
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In order for a force to qualify as having done work on an object, there must be a
displacement and the force must cause the displacement.
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Work examples
 A teacher applies a force to a wall and becomes exhausted. (NO Work)
 A book falls off a table and free falls to the ground. (YES)
 A waiter carries a tray full of meals above his head by one arm straight across
the room at constant speed. (NO Work.)
 A rocket accelerates through space. (YES)
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The Mathematics of Work
Words: The work done on an object is equal to the product of the force
on the object and the displacement on an object
(WORK = FORCE × DISPLACEMENT)
Formula: W=Fd
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The meaning of negative work:
 On occasion, a force acts upon a moving object to hinder a displacement.
 Examples: 1. a car skidding to a stop on a roadway surface, or
2. a baseball runner sliding to a stop on the infield dirt.
 In such instances, the force acts in the direction opposite the objects motion in
order to slow it down.
 The force doesn't cause the displacement but rather hinders it. These situations
involve what is commonly called negative work.
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Units of Work: The unit that work is measured in is the Joule (J)
Equation: W = Fd
Units:
J = Nm
The Work–Energy Theorem and Power
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The Work – Energy Theorem says that the energy in a closed system is conserved
except for the work done.
Work can be positive or negative.
Positive work will add energy to the system and negative work will subtract energy
from the system.
The math of the Work-Energy Theorem
(this is for closed systems!)
TME = KE + GPE + SPE
TME initial ± W= TME final
KEi + GPEi + SPEi ± W = KEf + GPEf + SPEf
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Another way to look at work:
W = TME final – TME initial
Power
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Definition: Power is the rate at which work is done and is measured in Watts. It is
defined as the amount of work done (in Joules) in one second of time. One way to
think of Power is how fast energy is being lost or gained.
The mathematics of power
Equations:
P = W/t
And we know:
W=Fd
So Power is also equal to: P = F d /t = F v
Example Power Problems:

Power Example #1:
A box is pushed with a force of 15 N across a flat surface and moves a
distance of 8 m in a time of 6 s. How much power was generated moving the
box?
Wanted:
Given:
P
1. F = 15N
2. d = 8m
3. t = 6s
Equation(s): P = F d / t
Sub.:
P = (15N)(8m)/6s
Math:
= 120Nm/6s
= 20
Answer:
P = 20W

Power example #2:
A new conveyor system at the local packaging plan will utilize a motorpowered mechanical arm to exert an average force of 890 N to push large
crates a distance of 12 meters in 22 seconds. Determine the power output
required of such a motor.
Wanted:
Given:
P
1. F = 890N
2. d = 12m
3. t = 22s
Equation(s): P = F d / t
Sub.:
P = (890N)(12m)/22s
Math:
= 10,680Nm/22s
= 485 W
Answer:
P = 485 W