chapter 5 work and machines page 126-131. review motion – distance, time, speed (velocity),...
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Chapter 5 Work and Machines
Page 126-131
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Review• Motion– Distance, time, speed (velocity), acceleration– Speed = rate of change of distance – Acceleration = rate of change of velocity
• Force– Force, mass, acceleration– A force is required to change the velocity of an object.• F = ma
– Weight is the gravitational attraction on an object• W = mg (g= 9.8 m/s2 the acceleration of gravity)
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Review of Energy• Energy is the ability to do work.• Different forms of energy• Different types of mechanical energy– KE = ½ mv2, PE (GPE)= mgh
• Energy is always conserved• Energy may be conserved but it is often changed
into a useless form (heat) by friction and air resistance.
• But Einstein said, E = mc2
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Work• Work makes something move• Work is the transfer of energy when a force
makes an object move.• Two conditions– A force must make something move– The direction of motion must be in the direction of the
force– Carrying books
• Work is a transfer of energy• W= Fd (force in Newtons (kg m/s2, distance in m)
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Practice Problems P 128:1,2,3
• You push a refrigerator with a force of 100N. If you move the refrigerator a distance of 5m while you are pushing, how much work is done?
• Given: • Asked: ?units• Formula:
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Page 128 Problem 1 A couch is pushed with a force of 75N and moves a distance of 5m across the floor. How
much work is done in moving the couch?
• Given:• Asked:• Formula: W=Fd
• Substitute:
• Answer:
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Problem 2 A lawn mower is pushed with a force of 80N. If 12,00J of work is done in mowing the lawn, what is
the total distance the lawn mower was pushed?
• Given:• Asked:• Formula: W=Fd
• Substitute:
• Answer:
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Problem 3 The brakes on a car do 240,000J of work in stopping a car. If the car travels a distance of 50m while the brakes are being applied, what is the force the brakes exert
on the car?• Given:• Asked:• Formula: W=Fd
• Substitute:
• Answer:
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Problem 4 The force needed to lift an object is equal in size to the gravitational force on the object. How much work is done in lifting an object with a mass of 5.0 kg a
vertical distance of 2.0 m?
• Given:• Asked:• Formula: W=Fd
• Substitute:
• Answer:
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Power
• Power is the rate of doing work.• How much work is done in 1 sec• P = W/t• W (joules) joules = kg m2/s2
• Unit for power is joules/second called a Watt• A Watt is kg m2/s3
• Since work is energy transferred • P= E/t
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Power Example• You push a box with a force of 100 N across
the floor 5 meters it takes you 45 seconds. Your friend pushes a similar box also with a 100 N force also 5 meters across the floor but it only takes your friend 30 seconds.– How much work do you each do?
– How much power do you each expend?
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Problem 1 Page 130In lifting a baby from a crib, 50 J of work are done.
How much power is needed if the baby is lifted in 2 s?
– Given:
– Asked: Units?
– Formula: P= W/t
– Substitute:
– Answer
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Problem 2 Page 130If a runner’s power is 130 W as she runs, how much
work is done by the runner in 10 minutes?
– Given:
– Asked: Units?
– Formula: P= W/t
– Substitute:
– Answer
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Problem 3 Page 130The power produced by an electric motor is 500 W.
How long will it take the motor to do 10,000 J of work?
– Given:
– Asked: Units?
– Formula: P= W/t
– Substitute:
– Answer
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Energy and Work
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Power Practice Problems P130: 1,2,3• You do 900 J of work pushing a sofa. If it took 5s
to move the sofa, how much power did you use?– Given
– Asked? Units?
– Formula
– Substitute
– Answer
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Section Review P131: 5,6,7• Work is a transfer of energy• W= Fd • (force in Newtons (kg m/s2, distance in m)• N*m or Joules• Power is the rate of doing work.• How much work is done in 1 sec• P = W/t• W (joules) joules = kg m2/s2 , t= seconds• Joules/second = Watt
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Machines (P132)
• A machine is a device that makes work easier.• Simple machines (not motorized)• How machines make work easier– A machine can increase the force that is applied to an
object.- Car jack (p 132) Multiplies force– A machine can make work easier by increasing the
distance which reduces the force needed. Ramp (p133)– A machine can change the direction of the force-
splitting wedge (p133)
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Work Done by Machines
• Figure 7 Page 133• Choice – lift the chair straight up the height of
the truck.• W = Fg x height
• Or- slide the chair up the ramp• Work will be the same• Distance will be greater so force will be less
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Work, Distance and ForceMachines don’t decrease the work (real machines increase the work), but they can decrease the force required.
Ideal machine (p135) Workin= Workout
3 m5 m
4 m
100 N
Wout= 100 N * 3m = 300J
Win = Fin *5 m , Wout= Win Wout= 300J but d=5m so F= 60N
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Factors of Work• Input force – the force applied to the machine• Output force – the force applied by the
machine• Input distance – the distance the input force
moves• Output distance – the distance the output
force moves.• work in = input force x input distance• work out = output force x output distance
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Pulling a Nail
• work out = work in
• Figure 10 page 135• hammer claw moves 1 cm to pull a nail• Handle moves 5 cm• Output force of 1,500 N• Input force = ?
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Ideal Machines (P135)• Workin= Workout
Forcein X distancein= ForceoutX distanceout
• Fin X 5cm = 1500N X 1 cm
• Fin= ?
• distance units cancel out if the same unit
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Ideal Mechanical Advantage
IMA = F out
Fin
d in
dout
=
Fin X din= FoutX dout
F out
Fin
d in
dout
=
Workin= Workout
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Real Machines• Work out< Workin
• Some of the input work is always converted into heat by friction.• Efficiency tells us how much of the
input work is converted into output work.
efficiency = Workout
Workin x 100% < 100%
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Assignment
• Chapter 4 Practice Problems• Practice Problems P 128:1,2,3 • Power Practice Problems P130: 1,2,3• Section Review P131: 5,6,7• Page 137 – Applying math 5-6-7
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Simple MachinesSection 3
Page 138-146
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Types of Simple Machines
• Simple machine does work with only one movement.
• Lever• Pulley• Wheel and Axle• Inclined Plane
• Compound Machines
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Lever• A bar that is free to pivot (turn) about a fixed
Point• The fixed point is called the fulcrum• Input arm – distance from the input force to
the fulcrum• Output arm – The distance from the output
force to the fulcrum.• If output arm is shorter than the input arm
then the output force is greater than the input force.
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Types of Levers (P138-139)First-class Lever
Third-class Lever
Second-class Lever
Input Force
Input Force
Input Force
Output Force
Output Force
Output Force
Output Distance
Input Distance
Input Distance
Input Distance
Output Distance
Output Distance
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Ideal Mechanical Advantageof a Lever
IMA = L input L output
IMA = Force output Force input
Force output x Length output=Force input X Length input
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Pulley Types (P 141-142)
Fixed Pulley
Input Force 4NOutput Force 4N
4N
Movable Pulley
Output Force 4N
Input Force 2N
4N
2N
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Types of Pulleys (cont)
Block and Tackle
Output Force 4N
4N
1NInput Force 1N
1N
1N1N
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Ideal Mechanical Advantageof a Pulley
IMA = Number of Strings Lifting the Load
IMA = Force output Force input
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Wheel and Axle (P143)
Input Force
Output Force
rarw
IMA = Radius of wheel Radius of axel
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3 m
100 N
F= 100N, height = 3m W= 100 N * 3m = 300J
W= 300J but d=5m so F= 60N
Inclined Plane
IMA = Length of Slope Height of Slope
5 m
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Types of Inclined Plane
• Ramps, inclines, road grades• Screw• Wedge• IMA = length of slope/height of slope• Page 144
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Compound Machines
• Two or more simple machines that operate together.–Can opener–Automobile–Space shuttle
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Section 3 Review
• Page 146 Applying Math 5-6-7
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Assignment• Practice Problems P 128:1,2,3 • Power Practice Problems P130: 1,2,3• Section Review P131: 5,6,7• Page 137 – Applying math 5-6-7• Page 146 Applying Math 5-6-7• Practice Problem Page- Due Monday • Note Taking Worksheet- due Monday• Lab on Tuesday• Chapter 5 Review Page 152-153• 1-20, 24-28 Due Wednesday• Test on Chapter 5 Friday
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Formulae
W= F*d units N*m = Joule
P= W/t or P= E/t unit J/s = Watt (kW)
F = m*a , FG= m*g unit kg m/s2 = Newton
efficiency = Workout
Workin x 100% < 100%
IMA = F out
Fin
d in
dout
=
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IMA of Simple Machines
Inclined Plane
Lever
Wheel and Axle
Pulley
MA = Force output Force input
All Simple Machines
IMA = L input L output
IMA = Number of Strings Lifting the Load
IMA = Length of Slope Height of Slope
IMA = Radius of wheel Radius of axel
> 1
efficiency = Workout
Workin x 100% < 100%
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Chapter 5 ReviewLever Input Arm Output Arm IMA
A 25 75
B 53 42
C 36 36
D 32 99
E 10 30
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Problems Page 153
• Given:
• Asked: Units
• Formula:• Substitute:• Answer
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Friday November 7• Get out your calculators and Reference Sheets• Put away your books and notes.
• Answer all the questions on the test sheets• Don’t overlook problems on the back of the test.• When you have finished, place test in the tray.• Place all homework in the tray• After the test sit quietly and begin reading chapter 6
on page 158.