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Work and Energy
Energy
 Ability to do work
 Scalar quantity
Work
 Transfer of energy to an object
 When an object is displaced by a force
 W = F∆d
1joule = newton∙meter
= N∙m
 Scalar quantity
The work done on the object gives energy to that
object.
Example of work:
 Lifting a pig above your head from the floor
Example of no work:
 Holding a pig above your head in your front yard
and proclaiming to the world you are “King of
Pigs”
Comparison of work quantities:
 A short person does less work than a tall person
in lifting a pig over one’s head.
Lesson learned:
 Hire short people to lift pigs.
A crate is pushed along a floor with a force of 20N
for a distance of 30m. How much work is done in
pushing the crate?
F = 20N
d = 30m
W=?
W = F∆d
= (20N)(30m)
= 600 N∙m
= 600J
How much energy does the crate gain?
600J
A 10kg pumpkin is lifted a vertical distance of 5m.
How much work is done in lifting the pumpkin?
m = 10 kg
d = 5m
F = 98N
w = mg
= (10kg)(9.8m/s²)
w = 98N
W = F∆d
= (98N)(5m)
W = 490N∙m
= 490J
Power
 Amount of work done per unit of time
 How fast an object does work
 Rate of work performed
 Scalar quantity
 p = W = Fd = Fv
t = t
1 watt = joule/second = newton∙meter/second
= kg∙m²
s³
p=W
t
10J = 5 watts
2sec
10J = 10 watts
1sec
A hoist lifts a gorilla weighing 600N a distance of
10m in 5seconds. How much power does the hoist
deliver?
F = 600N
p = Fd = (600N)(10m)
d = 10m
t
5sec
t = 5sec
= 100 watts
p=?
How long will it take a 1800 watt motor to push a
banana boat a distance of 2000m with a force of
10N?
p = W = F∆d
t = t
1800watts = (10N)(2000m)
t
1800watts·t = 20,000 N·m
t = 11.1 sec
A 100N force exerted on a goober (snot ball) causes
the goober to move with a constant velocity of 5m/s.
How much power does the goober exhibit?
p = Fv
= (100N)(5m/s)
= 500J/s
= 500watts
A moment of reflection:
 1watt = 1J/sec (MKS)
 1kilowatt = 1000watts
 1hp = 550ft·lb/sec (British System)
 1hp = 746watts
Work, Energy, and Machines
Machines make work easier by multiply the force put
into a machine. More force delivered by the machine
allows less force applied by a person.
Simple Machines (2 or less parts)
 work input = work output
Complex Machines (more than 2 parts)
 work input > work output
 a portion of work in is used to overcome the
friction of the moving parts
Machine Efficiency
 Efficiency = Work output x 100%
Work input
Energy is the ability to do work
Work = Energy
Hammer to nail example
 Swinging down of a hammer generates Energy
 Hammer has the ability to do work
 Hammer drives a nail a distance with a force,
work is done
Energy
 Scalar quantity
 Joules is the unit
 Is measured by the amount of work it can do
Forms of Energy
Thermal Energy (heat)
 total kinetic energy possessed by individual
particles comprising an object
Internal Energy
 total potential energy (PE) and kinetic energy
(KE) possessed by individual particles
comprising an object. Excludes the PE and KE
of the object as a system
The PE and KE of water molecules in a phase
change (ice changing to water or to steam) is
different from the PE and KE of a chunk of ice in a
free fall
Nuclear Energy
 energy released by nuclear fission or fusion
Chemical
 Stored form of energy.
 Explosives, Fuels, Food, Oxidants
Thermal
 Heat
Electromagnetic Energy
 energy associated with electric and magnetic
fields
 gamma rays, x-rays, light, IR, radar, microwave,
radio, TV
Potential Energy
 energy possessed by an object due to its position
or condition (distortion)
A cat held 3 meters above a toilet
A stretched rubber band
A compressed spring
A compressed catapult loaded with a cat
Gravitational Potential Energy
 ∆PE = mg∆h
 ∆PE = work
 work done to lift a cat to a new height is equal
to its change in PE
 When the cat is dropped from its new height it
will hit the toilet bowl with a KE equal to the PE
it gained when it was lifted to its new height
 ∆PE = W = F∆d
When the cat is raised, a force is exerted on it
equal to its weight or against the earth’s gravity.
Work is done on the cat
The cat gains PE
All that work or PE is changed to KE allowing the
cat to hit the toilet bowl with same amount of
force it took to raise the cat.
A 10kg cat is raised a vertical distance of 5m.
What is the resulting change in the cat’s PE?
∆PE = mg∆h
= 10kg(9.8m/s²)(5m)
∆PE = 490kg·m²/s²
= 490J
When the cat is dropped from a height of 5m, the
cat will hit the floor with 490J of kinetic energy.
m = 10kg
g = 9.8m/s²
Devices for Converting Energy
 Photo Cell – light energy
electrical
energy
 Generator – mechanical energy electrical
energy
 Gasoline Engine – mechanical energy
Heat energy
Light Energy
Electrical Energy
Sound Energy
v
A
E
B
D
C
The PE of the cart is least at ________
The PE of the cart is greatest at ______
The KE of the cart is least at _______
The KE of the cart is greatest at ______
Kinetic Energy
 Moving energy
 Scalar quantity
 Unit = joule
 KE = ½ mv²
 1J = kg·m/s
8m/s
Oscar the Magician flings his 10kg rabbit over his
head at 8m/s. How much KE does the rabbit gain
from this action?
m = 10kg
v = 8m/s
KE = ?
KE = ½ mv²
= ½ (10kg)(8m/s)²
KE = 320J
If Oscar the Magician threw the rabbit over his
head with twice the speed, how much KE would
be gained by the rabbit?
KE = ½ mv²
= ½ (10kg)(16m/s)²
KE = 1280J
doubling the v
produces 4x the
energy
KE α v²
KE α 2² KE α 4
Work – Energy Relationship
W = ΔPE = mgh
The work done in raising a lizard result in an increase
in gravitational PE
Conservation of Energy
 ΔPE + ΔKE = 0
 Sum of KE, PE, and internal energy remains
constant
 Closed System – no external forces doing a work
on the system, no work being done by the
system, and no transfer of energy in and out of
the system.
Conservation of Energy Diagrams
Nonideal Mechanical Systems
 Friction acts on a system
 Friction converts some KE of a moving object
into Internal Energy
 Internal Energy is the PE or KE of the particles
(molecules) of the object which translates as a
temperature increase
 ET = PE + KE + Q, ET is total energy of a
noinideal system
30m
5kg
20N
B
10m
A
Determine the work done against gravity in moving
the box from A to B.
W = ΔPE = mgh
= (5kg)(9.8m/s²)(10m)
= 491J
Determine the work done against friction in moving
the box from A to B.
W = Fd
= (20N)(30m)
= 600J
Wfriction = 600J – 491J = 109J
Potential Energy of a Spring
 F = kx (Hooke’s Law)
k = spring constant
F = force
 PES = ½ kx²
x = distance
A spring with a spring constant of 40N/m is stretched
a distance of 1.5m. What force is required to keep it
stretched?
k = 40N/m
x = 1.5m
F= ?
F = kx
= (40N/m)(1.5m)
= 60N
What is the PEs of the stretched spring?
PES = ½ kx²
= ½ (40N/m)(1.5m)²
PES = 45N·m = 45J