Transcript WorkPowerEnergy

Chapter 5
Work, Power and Energy
Objectives

Section 1 – Work
• Recognize the difference between the scientific
and ordinary definitions of work
• Define work by relating it to force and
displacement
• Identify where work is being performed in a
variety of situations
• Calculate the net work done when many forces
are applied to an object
Objectives

Section 2 – Energy
• Identify several forms of energy
• Calculate kinetic energy for an object
• Apply in the work-kinetic energy theorem to solve
problems
• Distinguish between kinetic and potential energy
• Classify different types of potential energy
• Calculate the potential energy associated with an
object’s position
Objectives

Section 3 – Conservation of Energy
• Identify situations in which conservation
of mechanical energy is valid
• Recognize the forms that conserved
energy can take
• Solve problems using conservation of
mechanical energy
Objectives

Section 4 – Power
• Relate the concepts of energy, time and
power
• Calculate power in two different ways
• Explain the effect of machines on work
and power
Work
Work – the product of force and
displacement, as long as they are
both in the same direction.
W  F d
This is a vector times a vector.
In this case, a vector times a vector gives a
scalar.
Work is measured in Joules (J) which is the same
as a Newton x meter.
Example
How much work is done on an object if
a force of 30 Newtons displaces the
object 200 meters?
W  F d
W  (30 N )(200m)
W  6000 J
What happens if the force and displacement
are not in the same direction?
F

d
W  ( F  cos )  d
or
W  Fd cos
Where θ is the angle between the two
vectors
Note: We’ll talk about where the cosine magically comes from in class.
Example:
As Alex pulls his red wagon down the
sidewalk, the handle of the wagon
makes an angle of 60o with the
pavement. If Alex exerts a force of
100 Newtons along the direction of
the handle,
60 how much work is done
when the displacement of the wagon
is 20 meters along the ground?
o
20 m
60o
20 m
W  ( F cos )d
o
W  (100 N )(cos60 )(20m)
W  1000 J
Example:
A constant force of 50 Newtons is
applied over a distance of 10 meters.
Draw a graph to represent this
information. Calculate the work
done by this force. Calculate the
F (N)
W

F

d
A

b

h
area
under
the
graph.
How
do
they
50
W  (50 N )(10m) A  (10m)(50 N )
compare?
W  500 J
A  500 J
10
d (m)
If given a graph of
Force vs. Displacement
The work can be found by calculating
the area under the graph.
That’s enough for now…
Lets assume you own a moving
company that employs two workers.
Employee one moves ten boxes and
gets the job done by lunch.
Employee two moves ten boxes but
takes his time and finishes by 5pm.
Who does more work?
Who would you pay more?
While there is no correct answer here,
there is an added piece that work
alone does not address.
Time is the added piece.
Power
Power is the rate at which work is
done.
W
P
t
Unit:
Joule
 Watt
sec
Example:
If 3000 Joules of work is performed on
an object in 1.0 minute, what is the
power expended on the object?
W
P
t
3000 J
P
60s
P  50W
Example:
A 200 Newton force is applied to an
object that moves in the direction of
the force. If the object travels with a
constant velocity of 10 m/s, calculate
the power expended on the object.
W F d
 F v
P 
t
t
P  (200 N )(10 ms )
P  2000W
Energy
When work is done on an object, the
energy of the object is changed.
Kinetic Energy
F  ma
W  Fd
W  ma  d
v  v  2ad
2
2
2
1
v22  v12
ad 
2
2
2
 v2  v1 
W  m

 2 
W  mv  mv
1
2
2
2
1
2
2
1
The work done on the object changes
a quantity called kinetic energy that
is related to the mass and the square
of the speed of the object.
W  mv  mv
1
2
2
2
KE  mv
1
2
2
1
1
2
2
W  KE2  KE1  KE
This is known as the work-energy theorem.
Example:
A 10kg object subjected to a 20 Newton
force moves across a horizontal
frictionless surface in the direction of the
force. Before the force was applied, the
speed of the object was 2.0 meters per
second. When the force is removed, the
object is traveling at 6.0 meters per
second. Calculate the following
quantities: (a) KE1 (b) KE2 (c) ΔKE (d) W
and (e) distance.
KE1  mv
m 2
1
KE1  2 (10kg )(2.0 s )
1
2
2
1
KE1  20 J
KE2  mv
m 2
1
KE2  2 (10kg )(6.0 s )
1
2
2
2
KE2  180 J
KE  KE2  KE1
KE  180 J  20 J
KE  160 J
W  KE
W  160 J
W  Fd
W
d
F
160 J
d
20 N
d  8.0m
That’s enough for today…
Gravitational Potential Energy
Lifting a book from the floor to the
table applies a force over a distance
therefore work is done on the book
but the kinetic energy is unchanged.
In this case the force of gravity is
overcome by work.
Fg  mg
W  F d
W  mg  d
PE  mg h
Example:
A 2.0 kilogram mass is lifted to a
height of 10 meters above the
surface of the Earth. Calculate the
change in the gravitational potential
energy of the object.
PE  mg h
PE  (2.0kg )(9.8 sm2 )(10m)
PE  196 J
Conservation of Energy
Energy can not be created or destroyed.
In the system, the sum of the potential
energy and kinetic energy will be
conserved; a change in one is
accompanied by an opposite change in
the other.
Example:
A 2.0 kg ball is tossed with a velocity
of 4.0 m/s straight into the air. How
high does the ballv 2travel
upwards?
2
2  v1  2ad
Calculate
the
m initial2 kinetic energy of
Givens: v  4.0
v1  2ad
s
1
the ball.
Calculate
the potential
m
2
a  9.8
2

v
s
1
energy of the
ball atd it’s
peak.

m
vtop  0 s
2a
m2
m  2.0kg
16 s2
d
2(9.8 sm2 )
d  0.815m
PE2  mgh
KE1  mv
2
1
1
2
m
PE

(2.0
kg
)(9.8
)(0.815m)
KE1  (2.0kg )(4.0 )
2
s2
m 2
s
1
2
KE1  16 J
PE2  16 J
KE2  0 J
PE1  0 J
PE  KE
or
E1  E2
PE1  KE1  PE2  KE2
m=1.0 kg
v1
Roller
Coaster
= 4.0m/s
h1 = 10 m
h1 = 5 m
h1 = 4 m
h1 = 2 m
Calculate the total energy and the velocity at the top of each hill.
v1 = 4.0m/s
m=1.0 kg
h1 = 10 m
h1 = 5 m
h1 = 4 m
h1 = 2 m
EETTT  KE
KEPE
PE
22
11
2 
E
E


mv
mv
mgh
mgh
2mv 
ETTT  122KE
PE
mgh
22
11
2 
1
mv
mv

E
E
mgh
mgh
2
1E
22T  2 mv TTmgh
2 mv  ET  mgh
22EE2
m
1
v
v


gh
gh
ET  2 (1kgv)(4
kg )(9.8 sm2 )(10m)
 2smmmE) 
2
22(1
gh
2(106JJ))
mm
vv  2(106


2(9.8
2(9.8
)(2.0m
m))
2(106
J
)
22)(4.0
11kg
kg E  106smJ
s
v  1kg T 2(9.8 s2 )(5.0m)
mm m
vv v11.6
13.2
 10.7
ss s
Pendulum
Simple Harmonic Motion – In the
absence of friction the pendulum will
swing back and forth continuously.
Example:
A pendulum whose bob weighs 12
Newtons is lifted a vertical height of
0.40 meter from its equilibrium
 PE
PE  mg Calculate
h
position.
theKE
change
in
PEpotential
 (12 N )(0.40
m) between
KE maximum
(4.8 J )
energy

PE  4.8
J equilibrium
height,
KE  4.8the
J
2
height
and
1
KE  2 mv
gain in kinetic energy,
and the
2 KE
v m
velocity at the equilibrium point.
2(4.8 J )
v  1.2 kg
v  2.8 ms