Transcript Chapter 10: Energy, Work and Simple Machines

Energy, Work and
Simple Machines
Chapter 10
Physics
Objectives: The student will be
able to:
1. distinguish between work in the scientific sense as compared
to the colloquial sense.
2. write the definition of work in terms of force and displacement
and calculate the work done by a constant force when the force
and displacement vectors are at an angle.
3. state and apply the relationship that work done with no
opposing force equals the change in kinetic energy.
4.state and apply the relationship that work done against gravity
equals the change in gravitational potential energy.
5.Define and calculate power from calculating the amount of
work done by an object.
What is work?
In science, the word work has a
different meaning than you may be
familiar with.
The scientific definition of work is: using
a force to move an object a distance
(when both the force and the motion of
the object are in the same direction.)
Work
 Work has its own meaning in physics.
 Work is done on an object when an
applied force acting on the object
moves the object over a distance.
 Work depends on two factors.
 Force (F)
 Displacement (d)
Work
Work = Force x Displacement
W = Fd
Unit for Work = Newton Meter (Nm)
1 Nm = 1 Joule (J) (Same as Energy)
Work is a scalar quantity (no direction)
In doing Work the Displacement has to
be in the same direction as the Force!
Work or Not?
According to the scientific definition,
what is work and what is not?
A teacher lecturing to her class
A mouse pushing a piece of cheese with
its nose across the floor
7
What’s work?
A scientist delivers a speech to an audience of his
peers.
 No
 A body builder lifts 350 pounds above his head.
 Yes
 A mother carries her baby from room to room.
 No





A father pushes a baby in a carriage.
Yes
A woman carries a 20 km grocery bag to her car?
No
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Physicist’s definition of “work”
dist∥
dist
Work = F
x
dist∥
Scalar Dot Product?
A product is obviously a result of multiplying 2 numbers. A scalar is a
quantity with NO DIRECTION. So basically Work is found by
multiplying the Force times the displacement and result is ENERGY,
which has no direction associated with it.
 

W  F  x  Fx cos
A dot product is basically a CONSTRAINT
on the formula. In this case it means that
F and x MUST be parallel. To ensure that
they are parallel we add the cosine on the
end.
FORCE
Displacement
 

W  F  x  Fx cos 
  0 ; cos 0  1

W  Fx
Work
 

W  F  x  Fx cos
FORCE
Displacement
 

W  F  x  Fx cos 
  180 ; cos180  1

 
W   Ff x
Work Done by a Constant Force
The work done by a constant force is defined as the distance moved
multiplied by the component of the force in the direction of
displacement:
Work
 

W  F  x  Fx cos
In the figure above, we see the woman applying a force at an angle
theta. Only the HORIZONTAL COMPONENT actually causes the
box to move and thus imparts energy to the box. The vertical
component (FsinQ) does NO work on the box because it is NOT
parallel to the displacement.
Work
•
Relates force to change in energy
•
Scalar quantity
Independent of time
•
Atlas holds up the Earth
But he doesn’t move,
dist∥ = 0
Work= Fx dist∥ = 0
He doesn’t do any work!
Work
 Work can be Zero (WNET = 0) in three
ways;
 d = 0 (does not move or finishes
where it starts)
 FNET = 0 (v = 0 or v = constant)
 FNET is perpendicular to d (F | d)
Work
If F ll d, then W = Fd or W = -Fd
W = Fd, Force in same direction as
displacement
( = 0o: cos = 1, Positive Work)
W = -Fd, Force is in the opposite
direction as the displacement
( = 180o: cos = -1, Negative Work)
Work From a Force vs
Displacement Graph
If you have a Force vs Displacement
Graph, where the Force is in Newtons
(N) and the Displacement is in Meters
(m), you can find the Work by finding
the Area Under the Curve!
W = Area under F vs d graph
Formula for Work
Work = Force x Distance
 The unit of force is newtons
 The unit of distance is meters
 The unit of work is newton-meters
 One newton-meter is equal to one joule
 So, the unit of work is a joule
Work Done by a Constant Force
In the SI system, the units of work are joules:
SI unit = Joule
1 J = 1 Nm = 1 kgm2/s2
As long as this person does
not lift or lower the bag of
groceries, he is doing no work
on it. The force he exerts has
no component in the direction
of motion.
Work
 

W  F  x  Fx cos
FORCE
Displacement
 

W  F  x  Fx cos 
  90 ; cos 90  0

W  0J
Work can be positive or negative
• 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
Work performed climbing
stairs
 Work = Fd
 Force
 Subject weight
 From mass, ie 65 kg
 Displacement
 Height of each step
 Typical 8 inches (20cm)
 Work per step
 650N x 0.2 m = 130.0 Nm
 Multiply by the number of steps
Power
The rate of doing work
 Work = Fd
Power  Work / time
Power  Fd / t
Power  Force  velocity
Units: Fd/s = J/s = watt
Energy
Energy is the property that describes an
object’s ability to change itself or the
environment around it.
Energy can be found in many forms.
Kinetic Energy (KE) – energy of motion.
Potential Energy (PE) – energy gained
by a change in position or structure
Kinetic Energy (KE)
Moving objects possess Kinetic Energy.
KE = ½ mv2
Energy is a scalar quantity and has the
unit of Joule (J) (1 J = 1 Nm)
Hewitt – Work and PE
Work done against gravity
Mass (g)
Work
(joules)
W = mgh
Height object raised (m)
Gravity (m/sec2)
The formula for kinetic energy

A force (F) is applied to mass (m) and
creates acceleration (a).
 After a distance (d), the ball has reached speed (v),
therefore the work done is its mass times acceleration time
distance:
– W= fd = (ma) x d = mad
– Also: d = ½ at2

Replace d in the equation for work, combine similar terms:
– W= ma (½ at2) = ½ ma2t2
– Also: v = at, so v2 = a2t2

Replace a2t2 by v2 shows that the resulting work is the
formula for kinetic energy:
– W = ½ mv2
Period 3
starts here
Hewitt – PE and KE
Work-Kinetic Energy Theorem
Work-kinetic Energy Theorem:
•
Net work done on a particle equals the
change in its kinetic energy (KE)
W = ΔKE
W  KEf  KEo  mv  mv
1
2
2
f
1
2
2
o
Practice Questions
Explain who is doing more work and
why:
A bricklayer carrying bricks and placing
them on the wall of a building being
constructed
Or a project supervisor observing and
recording the progress of the workers from
an observation booth.
Practice Question
 Explain who is doing more work and why:
 A bricklayer carrying bricks and placing them on the wall of a
building being constructed
 Or a project supervisor observing and recording the progress
of the workers from an observation booth.
 Work is defined as a force applied to an object, moving that
object a distance in the direction of the applied force. The
bricklayer is doing more work.
Practice Question
 How much work is done in pushing an object 7.0 m
across a floor with a force of 50 N and then pushing it
back to its original position? How much power is
used if this work is done in 20 seconds?
Practice Question
 How much work is done in pushing an object 7.0 m
across a floor with a force of 50 N and then pushing it
back to its original position? How much power is
used if this work is done in 20 seconds?
Work = 7.0m x 50N x 2 = 700 Nm or 700 J
Power = 700J / 20s = 35 W
Practice Question
 How much power will it take to move a 10 kg mass at
an acceleration of 2 m/s2 a distance of 10 meters in 5
seconds? This problem requires you to use the
formulas for force, work, and power all in the correct
order.
Practice Question
 How much power will it take to move a 10 kg mass at
an acceleration of 2 m/s2 a distance of 10 meters in 5
seconds? This problem requires you to use the
formulas for force, work, and power all in the correct
order.
 F = ma
 F = 10 kg x 2 m/s = 20 N
 W = Fd
 W = 20 N x 10 m = 200 J
 P = W/t
 P = 200 J / 5 s = 40 watts
Practice Question
Calculate work done against gravity
A crane lifts a steel beam with a mass
of 1,500 kg. Calculate how much
work is done against gravity if the
beam is lifted 50 meters in the air.
How much time does it take to lift the
beam if the motor of the crane can do
10,000 joules of work per second?
Practice Question
Calculate work done against gravity
A crane lifts a steel beam with a mass of 1,500 kg.
Calculate how much work is done against gravity if
the beam is lifted 50 meters in the air.
How much time does it take to lift the beam if the
motor of the crane can do 10,000 joules of work per
second?
1.
2.
3.
4.
5.
You are asked for the work and time it takes to do
work.
You are given mass, height, and work done per
second.
Use: W = mgh.
Solve: W = (1,500 kg) ( 9.8 N/kg) (50 m) = 735,000 J
At a rate of 10,000 J/s, it takes 73.5 s to lift the beam.
Practice Question
Calculating kinetic energy
A car with a mass of 1,300 kg is going straight ahead at a speed of
30 m/s (67 mph). The brakes can supply a force of 9,500 N.
Calculate:
a) The kinetic energy of the car.
b) The distance it takes to stop.
Practice Question
Calculating kinetic energy
A car with a mass of 1,300 kg is going straight ahead at a speed of
30 m/s (67 mph). The brakes can supply a force of 9,500 N.
Calculate:
a) The kinetic energy of the car.
b) The distance it takes to stop.
1.
2.
3.
4.
–
–
You are asked for kinetic energy and stopping
distance
You are given mass, speed and force of brakes.
Use Ek = 1/2mv2 and W= fd
Solve for Ek = ½ (1,300 kg) ( 30 m/s)2 = 585,000 J
To stop the car, work done by brakes = Ek of car, so W = Ek
Solve for distance = W ÷ f = 585,000J ÷ 9,500 N = 62 m
Elaboration
Force, Distance, and Work – Trans 10
 Work and Energy Internet Activity
 P. 261 #s1, 2, 3
 P. 262 #s 4, 5, 6, 7, 8
 P. 264 # 10 and 12
 Questions and Problems
 Work and Power
 Stair Climbing and Power lab
 Section Review #s 15-21

Closure

Kahoot 10.1 Work and Energy