Download Work, Power, and Energy - Polson 7-8

Chapter 5:
Work,
Power,
and Energy
5.1 Objectives
• Understand the concept of work.
• Be able to perform work calculations.
Work
work: a force applied through a distance (not a
displacement!) Work is a scalar.
force (F)
f
distance (d)
W = (F·cosf)·d
units: N·m = J (joules)
James Joule: studied
the relationship between
work and thermal energy
Work
• positive (+) work is done if f < 90o
• zero (0) work if f = 90o
• negative (-) work if f > 90o
• net work: SW = (SF·cosf)·d
Work Problem
A 107 N golf bag is dragged 125 meters (at constant
speed) with a force of 8.5 N. The force is oriented 32o
above the horizontal. How much work is done by the
golfer? By friction? By gravity? By the normal force?
W = F·cosf·d
5.4 Power
power: the rate at which work is done (work ÷ time)
W
P
t
units: J/s = W (watts)
W in an equation is work
James Watt: inventor
W as units are Watts
of the steam engine
746 W = 1 hp
Power Problem
A weightlifter lifts a 275 kg mass
from the floor to a height of 1.92
m in only 1.48 seconds. How
much work is done by the
weightlifter? How much power is
used?
5.2 Objectives
• Understand the concept of kinetic energy (KE)
and the closely-related work-energy theorem
• Solve KE and work-energy theorem problems.
Kinetic Energy
A moving object is capable of doing work if it runs
into something else—it has stored energy.
kinetic energy: the energy held by a moving object
(due to relative motion)
F = m·a
2
KE = ½·m·v
F·d = m·a·d
Why?
W = m·a·d
force
W = m·a·½·a·t2
distance
W = ½·m·a2·t2
W = ½·m·v2
KE Problem
A major league baseball has a mass of 0.145 kg.
How much KE does a baseball have if it is
thrown at 44.4 m/s (100 mph)?
Work-Energy Theorem
If the sum of all the work done on an object
(by all the forces) is calculated, then the SW
will equal the change in kinetic energy of the
object.
SW = SF · d = DKE = KEf - KEi
SW = SF · d = ½ m·Dv2
therefore, d ~ Dv2
Work-Energy Theorem Problem
• Brakes apply FFK (= SF)
• How much SW is done to stop a 1450 kg car
traveling at 20 m/s and 40 m/s? [45 mph & 80
mph]
• What is the stopping distance in each case if
the brakes apply 7.5 kN of force?
Stopping Distance Chart
5.2 Objectives
• Understand the concept of potential energy,
particularly gravitational potential energy
(GPE).
• Be able to solve gravitational potential energy
problems.
Potential Energy
Something is required to do work. That “something”
is called energy.
potential energy: stored energy (due to the
presence of a force)
gravitational potential energy (GPE)…
W = F·d·cos f
W = FW·h·cos(0o) = FW·h
height
W = m·g·h
(h)
GPE = m·g·h
units: N·m = J
GPE Problem
A 43 kg television is moved from the bottom to the
top of a flight of stairs that is 2.62 m high. The stairs
angle upward at 45o. How much work is done? How
much GPE does the TV have at the top of the stairs?
5.3 Objectives
• Understand the law of conservation of energy.
• Use the law of conservation of energy to solve
assorted dynamics problems.
Conservation of Energy
Mechanical energy (ME) is the sum of KE and all PE.
law of conservation of energy: energy is conserved
when converted from one form to another (the
total ME remains constant).
SPEi + SKEi = SPEf + SKEf
Why?
vf2 = vi2 + 2·a·d
GPE
vf2 = 2·g·h
KE = ½·m·v2
height (h) KE = ½·m·2·g·h
KE = ?
KE = m·g·h
(= GPE = m·g·h)
Conservation of Energy Problem
With what speed must a ball be thrown directly
upward to reach a final height of 28 meters?
Conservation of Energy Problem
A roller coaster traveling at 16 m/s drops down a steep
incline that is 25 meters high and then moves up another
incline. What is the height of the second incline if the
roller coaster is moving at 12 m/s at its crest? Assume
the effect of friction is negligible.
website
Bullseye Lab
h1
razor
h2
dx = ?
What is the equation
to find the range?
dx = ?
It is an extremely
simple equation!
Objectives
• Be able to identify simple machines.
• Be able to explain how simple machines make
doing work “easier.”
• Be able to calculate the ideal mechanical
advantage (IMA), actual mechanical (AMA)
advantage, input work (WI), output work
(WO), and efficiency (e) of a simple machine.
Simple Machines
4 kinds: lever, inclined plane, pulley, wheel and axle
Simple machines generally make doing work easier
by reducing applied force (but distance is increased).
input work:
WA = FA·dA
output work:
WO = FO·dO
If no friction: WA = WO
If friction is present: WA > WO
Simple Machines
mechanical advantage (MA): factor by which
input force is multiplied by the machine
IMA  d A
dO
“ideal”
AMA  FO
FA
“actual”
efficiency: ratio of output work to input work
(indicates amount of friction in machine)
e  WO
WA
100