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Energy
Energy is anything that can be converted into work; i.e., anything that can
exert a force through a distance.
Energy is the ability to do work.
Basic Properties of Energy
1. Energy can be transferred from one
object or system to another.
2.
Energy comes in multiple forms.
3.
Energy can be converted from any one of
these forms into any other.
4.
Energy is never created anew or destroyed this is The Law of Conservation of Energy.
Mechanical Energy
•Energy associated with motion
Types of Mechanical Energy
Kinetic Energy
Kinetic Energy: Ability to do work by
virtue of motion. (Mass with velocity)
A speeding car
or a space rocket
• KE = ½ mass x velocity2
• KE = ½ mv2
Examples of Kinetic Energy
What is the kinetic energy of a 5-g bullet
traveling at 200 m/s?
5g
200 m/s
KE = 100 J
How fast must a 700 kg car drive in order
to have 78,750 J of kinetic energy?
K  12 mv 2  12 (1000 kg)(14.1 m/s) 2
1
𝐾 = 𝑚𝑣 2
2
1
78,750 𝐽 = (700 𝑘𝑔)𝑣 2
2
𝒗 = 𝟏𝟓 𝒎 𝒔
Types of Mechanical Energy
Potential Energy
Potential Energy: Ability to do work
by virtue of position or condition.
A suspended weight
Gravitational Potential Energy
• PEg = weight x height
•
PEg = [mass X gravitational acceleration] X height
• PEg = mgh
A stretched bow
Elastic Potential Energy
Examples of Potential Energy
What is the potential energy of a 50 kg
person in a skyscraper if he is 480 m
above the street below?
PE = mgh = (50 kg)(9.8 m/s2)(480 m)
PE = 235,200 J
A typical 747 airplane flying at an
altitude of 11 km has 2.7x1010 Joules
of gravitational potential energy.
What is the mass of this airplane?
PE = mgh
2.7x1010 J = (m)(9.8 m/s2)(11,000 m)
m = 250,464 kg
Heat Energy
• Energy from the internal
motion of particles of matter
The hotter something is,
the faster its molecules are
moving around and/or
vibrating, i.e. the more
energy the molecules have.
Chemical Energy
• The energy from bonds between atoms or ions
Electromagnetic Energy
•Energy of moving electric charges
Nuclear Energy
• Energy from the
nucleus of the atom
– Fusion is when two
atoms combine
• The Sun
– Fission is when the
atom splits
• Nuclear power plant
Mass-Energy Equivalence: E=mc2
Conservation
of Energy
Students will:
a) Identify situations on which
conservation of mechanical energy is
valid.
b) Recognize the forms that conserved
energy can take.
c) Solve problems using conservation of
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mechanical energy.
Mechanical Energy
• Mechanical Energy is the sum of
kinetic energy and all forms of
potential energy in a system.
• In the absence of “nonconservative”
resistive forces like friction and drag,
mechanical energy is conserved.
• When we say that something is
conserved, we mean that it remains
constant.
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THE PRINCIPLE OF CONSERVATION OF
MECHANICAL ENERGY
The Total Mechanical Energy (TME) of an
object remains constant as the object moves,
in the absence of friction.
TME  constant
TMEinitial  TME final
KEi  PEi  KE f  PE f
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3/4 Potential Energy, 1/4 Kinetic Energy
1/2 Potential Energy, 1/2 Kinetic Energy
1/4 Potential Energy, 3/4 Kinetic Energy
No Potential Energy, all Kinetic Energy
Conservation of Energy
All Potential Energy, no Kinetic Energy
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If friction and wind resistance are ignored, a bobsled run
illustrates how kinetic energy can be converted to potential
energy, while the total mechanical energy remains constant.
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Ski Jumping (no friction)
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Example 1: A person on top of a building throws a 4 kg ball
upward with an initial velocity of 17 m/s from a height of 30
meters. If the ball rises and then falls all the way to the ground,
what is its velocity just before it hits the ground?
17 m/s
m = 4 kg
vi = 17 m/s
vf = ?
g = 9.8 m/s
hi = 30 m
hf = 0 m
TMEinitial  12 mvi2  mghi
30 m
TME final  12 mv 2f  mgh f
continued on next slide
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Example 1 continued:
1
2
1
2
mv  mghi  mv  mghf
2
i
1
2
2
f
(4 kg)(17 m s )2  (4 kg)(9.8 m s 2 )(30 m)
 12 (4 kg)v 2f  (4 kg)(9.8 m s 2 )(0 m)
578 J  1176 J  12 (4 kg)v 2f  0 J
1754 J  12 (4 kg)v 2f
877
m2
s2
 v 2f
29.6 ms  v f
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Example 2: A 10 kg stone is dropped from a height of 6 meters
above the ground. Find the Potential Energy, Kinetic Energy,
and velocity of the stone when it is at a height of 2 meters.
At 6 m:
TME = PEi + KEi= mghi + ½ mvi2
= (10 kg)(9.8 m/s)(6 m) + ½ (10 kg)(0 m/s)2
= 588 J + 0 J
= 588 J
6m
2m
At 2 m:
TME = PEf + KEf = mghf + ½ mvf2
= (10 kg)(9.8 m/s)(2 m) + ½ (10 kg)(vf)2
= 196 J + ½ (10 kg)(vf)2
Therefore: 588 J = 196 J + ½ (10 kg)(vf)2
588 J – 196 J = ½ (10 kg)(vf)2
392 J = ½ (10 kg)(vf)2
solve for vf = 8.9 m/s
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Example 3: A Daredevil Motorcyclist
A motorcyclist (300 kg including the bike) is trying to leap across
the canyon by driving horizontally off a cliff with an initial speed of
38.0 m/s. Ignoring air resistance, find the speed with which the
cycle strikes the ground on the other side.
38.0 m/s
vf = ?
70 m
55 m
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Example 3 continued
1
2
1
2
mv  mghi  mv  mghf
2
i
1
2
2
f
(300 kg)(38 m s )2  (300 kg)(9.8 m s 2 )(70 m)
 12 (300 kg)v 2f  (300 kg)(9.8 m s 2 )(55 m)
216,600 J  205,800 J  12 (300 kg)v 2f  161,700 J
260,700 J  12 (300 kg)v 2f
1738
m2
s2
 v 2f
41.7 ms  v f
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•Example 4:
Starting from rest, a child on a sled zooms down a
frictionless slope from an initial height of 8.00 m. What is
his speed at the bottom of the slope? Assume he and
the sled have a total mass of 40.0 kg.
8.00 m
Given :
h  hi  8.00m
m  40.0kg
vi  0 ms
hf  0 m
Unknown :
v f  ??
40.0kg 0 ms 2  0 J
PEi  mghi  40.0kg 9.81 sm 8.00m   3139 J
KEi  12 mvi 
2
1
2
2
KE f  12 mv f  12 (40.0kg)v 2f
2


PE f  mgh f  40.0kg  9.81 sm2 0m   0 J
KEi  PEi  KE f  PE f
continued on next slide
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•Example 4 - continued
KEi  PEi  KE f  PE f
40.0kg v f
2
3139 J  20kg v f
0 J  3139 J 
1
2
2
0
3139 J
2
 vf
20kg
157
m2
s2
 vf
157
m2
s2
 vf
12.5 ms  v f
2
answer
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Example 5: Unknown Mass
A skier starts from rest and slides down the frictionless
slope as shown. What is the skier’s speed at the bottom?
start
m = unknown
vi = 0 m/s
vf = ?
g = 9.8 m/s
hi = 40 m
hf = 0 m
H=40 m
finish
L=250 m
continued on next slide
Example 5: Unknown Mass - continued
KEi  PEi  KE f  PE f
1
2
mvi2  mghi  12 mv2f  mghf
You can divide the mass out of the above equation.
1
2
1
2
mvi2  mghi  12 mv2f  mghf
(0 m s )2  (9.8 m s 2 )(40 m)  12 v 2f  (9.8 m s 2 )(0 m)
392
m2
s
2
28 ms
 12 v 2f
 vf
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•Example 6:
A ball is dropped from a height of 5 meters above the
ground. Using conservation of energy formulas, determine
the speed of the ball just before it hits the ground.