### Lesson 12

```Energy
 A central concept underlying all sciences
 However, concept of energy is very difficult to define
 Concept of energy is unknown to Isaac Newton. Its
existence was still debated in the 1850s
 A quantity that is often understood as the capacity to
do work on a physical system
Energy
 We observe only the effects of energy when
something is happening
 When energy is being transferred from one place to
another
 When energy is being transformed from one form to
another
Forms of Energy
 Thermal energy
 Elastic energy
 Chemical energy
 Sound Energy
 Electrical energy
 Mechanical energy
 Luminous energy
 Nuclear energy
 Magnetic energy
Work
 The amount of energy transferred by a force acting
through a distance
 Work = net force x distance
W = Fnet x d
 Unit of work: newton-meter (Nm) = joule (J)
Work
 A scalar quantity
 Independent of time
 Unit of work: newton-meter (Nm) = joule (J)
 1
J = 1 N x 1 m = 1 kgm2/s2
 1 KJ = 1,000 J
 1 MJ = 1,000,000 J
Work
Work
Work
Two ways to do work
 Force acts to change motion
 Force acts against another force
Work Example
 An 80 N force has been applied to a block and move
it 20 m along the direction of the force. How much
work has been done to the block?
Work Example
 An 80 N force has been applied to a block and move
it 20 m along the direction of the force. How much
work has been done to the block?
W=Fd
= 80 N x 20 m
= 1600 J
Work Exercise
 Calculate the work done when a 20-N force pushes a
cart 3.5 m ?
Work Exercise
 How much work is required to lift a 360 kilogram
piano to a window whose height is 10 meters from
the ground?
Work Exercise
 A box rests on a horizontal, frictionless surface. A girl
pushes on the box with a force of 18 N to the right
and a boy pushes on the box with a force of 12 N to
the left. The box moves 4.0 m to the right. Find the
work done by (a) the girl, (b) the boy, and (c) the net
force.
Work vs. Impulse
 impulse = net force x time
J=Ft
impulse changes the momentum of a system
unit of impulse: newton-second (Ns)
 work = net force x distance
W=Fd
work changes the energy of a system
unit of work: newton-meter (Nm) = joule (J)
Work and Force with an Angle
W = F cos θ d
d
Work Exercise
 If 100 N force has 30o angle pulling on 15 kg block for
5 m. What’s the work?
Work Exercise
 If 100 N force has 30o angle pulling on 15 kg block for
5 m. What’s the work?
W = F d cos
W = 100 N (5 m) cos 30o = 433 J
Positive or Negative Work
 Man does positive work
lifting box
 Man does negative work
lowering box
 Gravity does positive work when
box lowers
 Gravity does negative work when
box is raised
Power
 Power – The rate at which work is done
 Power = (Work Done) / (Time Interval)
P = W/t
 Unit of power: joule per second , or Watt (W)
 1
 1
watt (W) = 1 J / 1 second
kilowatt (kW) = 1,000 watts
 1 megawatt (MW) = 1,000,000 watts
Power
 When W is in the equation, W means work; when W
is in the unit, W means Watt
 In metric system of units, automobiles are rated in
kilowatts; in the US, we rate engines in units of
horsepower (hp).
 1 hp = 0.75 kW
or
134 hp = 100 kW
Power Example
 Calculate the power expended when a 500 N barbell
is lifted 2.2 m in 2 s
Power Example
 Calculate the power expended when a 500 N barbell
is lifted 2.2 m in 2 s
W = Fd/t = 500 N x 2.2 m / 2 s
= 1,100 J / 2 s
= 550 watts
Power Exercise
 An escalator is used to move 20 passengers every
minute from the first floor of a department store to
the second. The second floor is located 5-meters
above the first floor. The average passenger's mass is
60 kg. Determine the power requirement of the
escalator in order to move this number of passengers
in this amount of time
Power
 Power = Force x Average Velocity
P = W/t = Fd/t = Fv
Power Example
 A 500N force is applied to an object. If the object
travels with a constant velocity of 20 meters per
second, calculate the power expended on the object
Power Example
 A 500N force is applied to an object. If the object
travels with a constant velocity of 20 meters per
second, calculate the power expended on the object
P = Fv
= (500N)(20m/s)
= 10,000 W
= 10 kW
Power Exercise
 An elevator must lift 1000 kg a distance of 100 m at a
velocity of 4 m/s. What is the average power the
elevator exerts during this trip?
Mechanical Energy
 A compression of atoms in the material of an object,
a physical separation of attracting bodies, or a
rearrangement of electric charges in the molecules of
a substance
 Energy enables an object to do work
 Unit of energy: energy is measured in joules
Mechanical Energy
 Two most common forms of mechanical energy:
potential energy (PE) and kinetic energy (KE)
 Mechanical energy can be in the form of either one,
or the sum of the two
 Potential energy— the energy due to the position of
something, or
 Kinetic energy— the energy due to the movement of
something
Potential Energy
 Definition: An object may store energy by virtue of its
position
 The energy that is held in readiness is called potential
energy (PE), because in the stored state it has the
potential for doing work
Potential Energy
 Examples of potential energy:
 A stretched/compressed spring
 A drawn bow
 A stretched rubber band
 The chemical energy in fuels (at submicroscopic level)
 Any substance that can do work through chemical action
possesses potential energy (such as fossil fuels, electric
batteries and food)
Gravitational Potential Energy
 Work is required to elevate objects against Earth’s
gravity
 The potential energy due to elevated positions is
called gravitational potential energy (GPE)
 The amount of gravitational potential energy
possessed by an elevated object is equal to the work
done against gravity in lifting it
Gravitational Potential Energy
 The upward force required while moving at constant
velocity is equal to the weight (mg) of the object
 Gravitational Potential Energy = Weight x Height
GPE = Fgh = mgh
or
PE = mgh
 The gravitation potential energy depends only on mg
and h, does not depend on the path
 Unit of gravitational potential energy: kg m2/s2 =
joule (J)
Potential Energy Example
 The floor of the basement of a house is 3.0 m below
ground level, and the floor of the attic is 5.0 m above
ground level. What is the change in potential energy if
an 2.0 kg object in the attic is brought to the
basement?
Potential Energy Example
 The floor of the basement of a house is 3.0 m below
ground level, and the floor of the attic is 5.0 m above
ground level. What is the change in potential energy if
an 2.0 kg object in the attic is brought to the
basement?
PE = mgh
= (2.0 kg)(9.8 m/s2)(5.0 m + 3.0 m)
= 156.8 J
Potential Energy Exercise
 A box has a mass of 5.8kg. The box is lifted from the
garage floor and placed on a shelf. If the box gains
145J of Potential Energy (Ep), how high is the shelf?
Potential Energy Exercise
 A man climbs on to a wall that is 3.6m high and gains
2268J of potential energy. What is the mass of the
man?
Spring Force
 Hooke’s Law: Hooke's Law says that the spring
restoration force (FS) due to a spring is proportional
to the length that the spring is stretched or
compressed, and acts in the opposite direction

FS = kx


k
x
is spring constant, and
change in spring length from the equilibrium position
 Unit of spring force: newton (N)
Hooke’s Law
 The more force that was put on materials the
more they extended
 With some materials they also extended in a
regular way e.g. if the force was doubled so did
the extension
 This was true as long as their elastic limit was not
exceeded
Hooke’s Law
Force vs. Extension
Elastic limit
 The material no longer shows elastic behavior (i.e.
force is removed)
 The material is permanently deformed i.e., is
larger or longer than originally
Spring Constant
 k is called the spring constant and is a measure of the
stiffness of the spring or material
 Unit of of k: N/m (newtons per meter)
 The higher the k the stiffer the spring, it needs a large
force to for a given extension
Spring Force Example
 A linear spring has a constant of 377.16 N/m. How
much force is necessary to stretch it 0.39 m?
Spring Force Example
 A linear spring has a constant of 377.16 N/m. How
much force is necessary to stretch it 0.39 m?
Fs = kx
= (377.16 N/m)(0.39 m)
= 147.09 N
Spring Force Exercise
 It requires a force of 594.55 N to stretch a certain
linear spring 0.15 m. What is the constant for this
spring?
Spring Potential Energy
 Spring Potential Energy: A spring may store energy
due to compression or stretching and is called spring
potential energy (PES)
 The potential energy stored in spring is equal to the
work done to the spring:
 PES = FSx = (1/2FS)x = (1/2kx)x = 1/2kx2
PES = 1/2kx2
 Unit of Spring Potential Energy: N m = joule (J)
Spring Potential Energy Example
 Vincent's mountain bike has a spring with a constant
of 64 N/m in the front-wheel suspension, and it
compressed 0.17m when she hit a bump. How much
energy does the front spring now store?
Spring Potential Energy Example
 Vincent's mountain bike has a spring with a constant
of 64 N/m in the front-wheel suspension, and it
compressed 0.17m when she hit a bump. How much
energy does the front spring now store?
PEs = (1/2) kx2
= (1/2)(64 N/m)(0.17 m)2
= 0.92 J
Spring Potential Energy Exercise
 A spring has a potential energy of 22.14 J and a
constant of 676.96 N/m. How far has it been
stretched?
Kinetic Energy
 If an object is moving, then it is capable of doing
work. It has energy of motion, or kinetic energy (KE)
 The kinetic energy of a moving object is equal to the
work required to bring it to that speed from the rest,
or the work the object can do while being brought to
rest
 The kinetic energy of an object depends on the mass
and speed of the object
Kinetic Energy
 Work = Net Force x Distance = Kinetic Energy

KE = W = Fd = mad = m(v2/(2d))d = 1/2mv2
KE = 1/2mv2
 Unit of kinetic energy: kg m2/s2 = joule (J)
Kinetic Energy Example
 A 3 kg ball is rolling 2 m/s. How much kinetic
energy does it have?
Kinetic Energy Example
 A 3 kg ball is rolling 2 m/s. How much kinetic
energy does it have?
KE = ½ mv2
= ½ (3 kg) (2 m/s)2
=6J
Kinetic Energy Exercise
 Determine the kinetic energy of a 500-kg roller coaster
car that is moving with a speed of 20 m/s
 If the roller coaster car were moving with twice
the speed, then what would be its new kinetic
energy?
Kinetic Energy Exercise
 Missy Diwater, the former platform diver for the
Ringling Brother's Circus, had a kinetic energy of
12000 J just prior to hitting the bucket of water. If
Missy's mass is 40 kg, then what is her speed?
Kinetic Energy
 If the speed of an object is doubled (the speed is
squared), its kinetic energy is quadrupled
 It takes four times the work to double the speed
 It takes four times of work to stop a double-speed
object
 Work-Energy Theorem: Work changes energy. If no
change in energy occurs, then no work is done.
W= E
Work-Energy Theorem Example
A skater of mass 60 kg has an initial velocity of 12 m/s.
He slides on ice where the frictional force is 36 N. How
far will the skater slide before he stops?
Work-Energy Theorem Example
A skater of mass 60 kg has an initial velocity of 12 m/s.
He slides on ice where the frictional force is 36 N. How
far will the skater slide before he stops?
W = E = KE
Fnet d = 1/2mv2
36 N x d = ½ (60 kg)(12 m/s)2
d = 120 m
Work Exercise
 When a small brass ball is dropped into soft clay, it
makes a dent. If the ball hits with twice the speed, the
dent of the clay is
a) more or less the same depth.
b) twice as deep.
c) three times as deep
d) four times as deep.
Conservation of Energy
 Law of conservation of energy: Energy cannot be
created or destroyed. It can be transformed from one
form into another, but the total amount of energy
never changes
 Total Energy at point A = Total Energy at point B

PEA + KEA = PEB + KEB
 PE = mgh
 KE = 1/2mv2
Conservation of Energy Example
 A diver of mass m drops from a
board 10.0 m above the water
surface. Find his speed 5.00 m
above the water surface
Conservation of Energy Example
 A diver of mass m drops from a
board 10.0 m above the water
surface. Find his speed 5.00 m
above the water surface
PE = KE
mgh = ½ mv2
m(9.81 m/s)(10 m – 5m) = ½ m v2
v = 9.9 m/s
Conservation of Energy Exercise
 A diver of mass m drops from a
board 10.0 m above the water
surface. Find his speed right above
the water surface
Conservation of Energy Exercise
A skier slides down the frictionless slope as shown.
What is the skier’s speed at the bottom?
start
H=40 m
finish
L=250 m
Conservation of Energy Exercise
Three identical balls are thrown
from the top of a building with the
same initial speed. Initially,
Ball 1 moves horizontally.
Ball 2 moves upward.
Ball 3 moves downward.
Neglecting air resistance, which ball
has the fastest speed when it hits
the ground?
A) Ball 1
B) Ball 2
C) Ball 3
D) All the same
Conservation of Energy Exercise
Tarzan swings from a vine whose length is
12 m. If Tarzan starts at an angle of 30
degrees with respect to the vertical and
has no initial speed, what is his speed at
the bottom of the arc?
```