Transcript document

Chapter 6. Energy and Work
Chapter Goal: To develop
an understanding of
mechanical energy; its
transfer to and from a system
through work, its
transformation within the
system, and its conservation.
Ch. 6 – Student Learning Objectives
• To begin developing a concept of energy—
what it is, how it is transformed, and how it is
transferred.
• To introduce the concepts of kinetic and
potential energy.
• To introduce work as an energy transfer
mechanism.
Energy transfers and transformations – How the
universe moves
Electromagnetic
Conservation of Mechanical Energy – The Big
Picture
Two basic types of mechanical
energy:
• Kinetic energy (K) is an energy of
motion.
• Gravitational potential energy U
(GPE in our text) is an energy of
position in a gravitational field.
Under some circumstances (e.g.
freefall) these two kinds of energy
can be transformed back and forth
without loss from the system.
Kinetic energy
of the system
can change to
gravitational
potential
energy (U) and
vice versa
Work – an energy transfer mechanism
• Work: The transfer of
energy into or out of a
system by application of a
force, F, that acts through
some displacement, s.
• The units of work and energy
are Joules (J) where 1 Joule
is equal to 1 Newton-meter.
• Both force and displacement
must be non-zero for work to
be done on the object.
Kinetic energy
of the system
can change to
gravitational
potential
energy (U) and
vice versa
Work – an energy transfer mechanism
• W > 0: The force
interaction between the
system and environment
causes the system to
increase either its speed or
height; therefore the
system’s mechanical
energy increases.
Conservation of mechanical
energy tells us that this
same force interaction will
cause the energy of the
environment to decrease.
Kinetic energy
of the system
can change to
gravitational
potential
energy (U) and
vice versa
Work – an energy transfer mechanism
• W < 0: The force
interaction between the
system and environment
causes the system to lose
either speed or height. In
this case, the mechanical
energy of the system
decreases.
• As you may already have
guessed, in this case, the
energy of the environment
increases.
Kinetic energy
of the system
can change to
gravitational
potential
energy (U) and
vice versa
Energy and Work: The overall picture
• Work is a scalar, even
though it is derived
from two vector
quantities
• Positive work results in
an energy gain and
negative work in an
energy loss.
• The negative sign does
not indicate left, right,
up, down, as it does
with a vector quantity, it
represents less.
Kinetic energy
of the system
can change to
gravitational
potential
energy (U) and
vice versa
Work done by a constant force
• Consider our system to consist of the suitcase. The rest
of the universe (and especially the woman) is the
environment. By pulling on the suitcase with a force, F,
the environment increases the energy, (specifically the
kinetic energy) of the suitcase.
• Now let’s calculate the work done by the pulling force
on the suitcase.
Work done by a constant force
W = |F| |s| cos θ
• F and s are, respectively, the magnitudes of the force and
displacement vectors,
• θ is the angle (not necessarily acute) between the force
vector and the displacement vector when they are tail to
tail.
• This is the work due to the pulling force only.
Work done on the barbell by the gym dude
Wgd = |Fgd| |s| cos θ
If the force and displacement vectors
are in the exact same direction
(parallel), θ = 0˚, cos θ = 1.
If the vectors are exactly opposite of
each other,
θ = 180˚, cos θ = -1.
Work is a scalar quantity. The
negative sign means the system
(barbell) has less energy, not its
direction.
In the bottom, picture, the bar bell is
losing height in the gravitational
field (GPE)
Work done by a constant Force
If the angle between the
force and the displacement
is less than 90°, the work
done will be positive and
the object will gain energy.
If the angle is more than
90 °, the work done will
be negative and the object
will lose energy.
Work has its ups and downs, but is it positive or negative?
A crane lowers a steel beam into
place at a construction site. The
beam moves with constant speed.
Consider the work Wg done by
gravity and the work WT done by
the tension in the cable. Which of
the following is correct and why?
A.
B.
C.
D.
E.
WG and WT are both zero.
WG is negative and WT is negative.
WG is negative and WT is positive.
WG is positive and WT is positive.
WG is positive and WT is negative.
T
FG
Δs
Numerical Examples – A tow truck
A tow truck drags a stalled car of mass 1000 kg a
distance of 500 m at constant velocity. The tow cable
makes an angle of 30˚ with the road. A frictional force
of 1386 N opposes the motion.
a. How much work is done by each of the external forces
acting on the car?
b. Determine the net work done on the car.
Numerical Example – A tow truck
Known
m=1000 kg
θ = 30 ˚
Δs = 500m
fk = 1386 N
T cos θ = 1386 N (constant velocity)
Find
WT , Wf ,WFG
Wn Wnet
n
fk
T
θ
FG
Δs
Numerical Example – A tow truck
T = 1386N / cos θ = 1600 N
WT = TΔs cos θ
WT = (1600 N)(500m) cos 30˚
WT = 6.93 x 105 Joules
WFG = Wn = 0 (cos 90°!)
Wf = (1386N)(500) cos 180˚
Wf = - 6.93 x 105 Joules
Wnet = 0 (car gains neither speed
nor height, so we would expect
that).
n
fk
T
θ
FG
Known
m=1000 kg
θ = 30 ˚
Δs = 500m
fk = 1386 N
Find
T, WT , Wf
WFG ,Wn Wnet
Δs
Numerical Example – #9 – elevator going down
An elevator has a mass of 6020 kg.
It travels 6.0 m down to the next
floor. The tension in the cable
exerts a force of 70,000 N on the
elevator.
a. How much work is done by
each of the external forces
acting on the elevator?
b. Determine the net work done on
the elevator.
T
FG
Δs
Numerical Example – #9 – elevator going down
T
WFG = FG Δs cos θ
WT = TΔs cos θ
Wnet = WT + WG
Known
m=6020 kg
Δs = 6.0 m
T = 70,000 N
FG
Δs
Numerical Example – #9 – elevator going down
T
WFG = FG Δs cos 0° = 3.54 x 105 J
WT = TΔs cos 180 ° = -4.20 x 105 J
Wnet = WT + WG = -6.60 x 104 J
FG
Known
m=6020 kg
Δs = 6.0 m
T = 70,000 N
Δs
The hand does work on the box
In the figures below, identical boxes are moving at the same
initial velocity to the right. The same magnitude force, F, is
applied to each box by an unseen hand, for the distance, d,
indicated in the figures.
Rank these situations in order of the work done on the box by
the hand while the box moves the indicated distance to the
right. Negative values of work rank lower than positive works.
The box does work on the hand
In the figures below, identical boxes of mass 10 kg are moving at
the same initial velocity to the right on a flat surface. The same
magnitude force, F, is applied by a hand to each box for the
distance, d, indicated in the figures.
Rank these situations in order of the work done by the box on the
hand exerting the force F while the box moves the indicated
distance to the right. Negative work rank lower than positive
work.
The Work-Energy Theorem (in its simplest
expression)
• Kinetic energy is energy of
motion: K = ½ mv2 .
• The work done on a system
adds kinetic energy to or
takes it away from the
system. The Work-Energy
Theorem expresses that as:
W = ΔK
We will consider gravitational
potential energy shortly.
Kinetic energy
of the system
can change to
gravitational
potential
energy (U) and
vice versa
Work-Kinetic Energy Theorem.
Expanding the statement W = ∆K:
|F| |s| cos θ = ½ mvf2 – ½ mv02
Numerical example- Work Energy
A 58.0 kg skier is coasting down
a 25° slope, which is 57m
long. She has an initial
velocity of 3.6 m/s. A kinetic
friction force of 70 N opposes
her motion. Using the WorkEnergy Theorem, find her
speed at the bottom of the hill.
Numerical example- Work Energy
• Instead of using Newton’s 2nd law
and then kinematics to solve this
problem, we will analyze it from
the perspective of work and energy.
• 1. Draw a free body diagram and
list knowns and quantities to find.
• 2. Find the work due to all the
external forces and solve for net
work.
• 3. Use the Work Energy Theorem
to solve:
Numerical Example
no work due to normal force
Wf = 70 N x cos 180° x 57 m = -3990 J
WG = 568.4N x [cos (65°) or sin (25°)] x 57m =
13682 J
Wnet = 9692 J
Work-Energy Theorem:
(½ mv2)0 + Wnet = (½ mv2)1
2(½ mv20 + Wnet)/m = v12
[(58kg)(13 m2/s2) + 19384J] / 58kg = v12
v1 = 18.6 m/s
speed is found in one step. No need to do a
Newton’s Law analysis and a kinematics
analysis.
65
knowns
find: v1
m= 58 kg
v0 = 3.6 m/s
x0 = 0 m
x1 = 57m
fk = 70N
θ (incline) = 25°
w = mg = 568 N
EOC #24
• A skier slides horizontally along the snow for a
distance of 21 m before coming to rest. The
coefficient of kinetic friction between skier
and snow is 0.050. Initially, how fast was the
skier going? (mass isn’t given).
EOC #24
W = ∆K:
|F| |s| cos θ = ½ mvf2 – ½ mv02
μmg s cos 180 = ½ m(vf2 –v02)
The term for mass drops out and
vf=0
μg s cos 180 = ½(–v02)
v0 = +/- 4.5m/s, with positive
being physically possible.
n
fk
FG
Δs
Known
Find
µk = 0.05
v0
Δs = 21 m
fk = μk |n| = μmg
(on level ground)
fk is the only force
that does work
The Work-Energy Theorem, a Statement of the
Conservation of Energy Within a System
W = KEf - KE0
• The kinetic energy of a system
cannot be created or
destroyed; it can only be
transferred in or out (using
work in our studies) or
transformed into another kind
of energy.
• What are those other kinds of
energy?
• To answer that, let’s take a
closer look at the forces that
do work.
Kinetic energy
of the system
can change to
gravitational
potential
energy (U) and
vice versa
Conservative vs Non-conservative Forces
• conservative force – a force is conservative
when the work it does on a moving object does
not depend on the path between the object’s
initial and final position.
– the only conservative forces studied this class
is the gravitational force
– non-conservative forces include friction,
tension, propulsion forces, air resistance (in
fact, everything else!).
The Work-Energy Theorem, a Statement of the
Conservation of Energy Within a System
• Wc - the work done by conservative forces
• Wnc - the work done by non-conservative
forces
Rewrite the Work Energy Theorem as:
Wc + Wnc = KEf - KE0
Since FG is the only conservative force we will
study, Wc is the work done by gravity:
WG + Wnc = KEf - KE0
Work of Gravity
WFG = mg (Δs) cos θ
Using right angle
geometry and the
definition of a
conservative force, it
can be shown that
(Δs) cos θ will always be
equal to Δh, the height
above an arbitrary
origin.
Work of Gravity
WFG = mg (Δs) cos θ
can be written as
WFG = mg (Δh) cos θ
Note that this value will be
negative when the ball
goes up; positive when
the ball goes down.
Why?
Gravitational Potential Energy
Definition: The change in
gravitational potential
energy is numerically equal
and opposite in sign to the
conservative work done by
the force of gravity:
ΔPE = -WFG = mg (Δh)
Therefore: PE = mgh
Note: most texts use the letter U to
denote potential energy – and
other types of energy as well
Wnc + Wc = ΔK – (first version of the Work
Energy Theorem)
Wnc = ΔK – Wc but Wc = -ΔPE , so
Wnc = ΔK + ΔPE
• Any change in either the kinetic energy or the
potential energy of an object occurs because
work (either positive or negative) is done on
the object.
• If no work is done on the object, any change in
one type of energy must be balanced by a
corresponding change in the other type.
Work Energy Theorem
Wnc = ΔK + ΔPE can be written as:
Wnc = K - K0 + PE - PE0
• I like to rearrange this so it’s in “chronological
order”: K0 + PE0 + Wnc = K + PE or
½ mv02 + mgh0 + |F||s| cos θ = ½ mv2 + mgh
• Initial mechanical energy plus work done equals
final mechanical energy
• The work due to gravity has been replaced by
the change in PE, so don’t use both in one
problem: WG = - ΔPE
Conservation of Mechanical Energy
If the Wnc = 0 then the mechanical energy of the
object will not change and therefore is
conserved:
K0 + PE0 = K + PE
If the Wnc ≠ 0, then mechanical energy is not
conserved for the object (it just got transferred
to/from another object)
Rank, in order from greatest to least, the magnitude of change in Ug
(gravitational potential energy) of the sliding blocks, from top to
bottom of frictionless inclined surface (note all these cases represent
a loss in potential energy:
Reasoning Strategy - Work-Energy problems
• Draw a free body diagram and use Newton’s
2nd Law to determine the value of external
forces doing non-conservative work.
• Draw a modified pictorial representation with
an origin to identify values for K, PE, if
necessary.
• List knowns and quantities to find.
• Use the Work Energy Theorem modified for
the problem
• Recall h = ∆s sin θ for inclined slopes
EOC # 35
A 55.0 kg skateboarder starts out with a speed of
1.80 m/s. He does +80.0 J of work on himself
by pushing with his feet against the ground. In
addition, friction does -265 J of work on him.
In both cases, the forces doing the work are
nonconservative. The final speed of the
skateboarder is 6.00 m/s.
a. Calculate ΔPE
b. how much has the vertical height changed?
EOC #83
A basketball of mass 0.60 kg is dropped from rest from a height
of 1.05 m. It rebounds to a height of 0.57 m.
(a) How much mechanical energy was lost during the collision
with the floor?
(b) A basketball player dribbles the ball from a height of 1.05 m
by exerting a constant downward force on it for a distance of
0.080 m. In dribbling, the player compensates for the
mechanical energy lost during each bounce. If the ball now
returns to a height of 1.05 m, what is the magnitude of the
force?
6.7 Power
•Power is the rate at which energy is transformed or
transferred:
E
P
t
•Work is an energy transfer mechanism. The work done
on an object results in a change in its energy:
W = ΔE
•Therefore we can say:
Work done
W
P

Time interval t
joule s  watt (W)
Example
A man pushed a crate 10 m with a 10 N force in
the direction of motion. In which case did he
use the most power?
A. He pushed it for 1 s
B. He pushed it for 10s
C. He pushed it for 0.1 s
D. The amount of time doesn’t matter; they are
all the same power.
Another darn ratio problem (FC#25)
The power needed to accelerate a projectile from rest to
its launch speed v in a time t is P. By what factor does
P change if the same projectile is accelerated from
rest to a launch speed of 2v in a time of ½ t? Assume
no change in height during this time.
a. Increases by a factor of 2
b. Increases by a factor 4
c. Increases by a factor of 8
d. Stays the same; increase in v offset by decrease in
time.
Another darn ratio problem (FC#25)
The power needed to accelerate a projectile from rest to its launch
speed v in a time t is P. By what factor does P change if the
same projectile is accelerated from rest to a launch speed of 2v
in a time of ½ t? Assume no change in height.
a. Increases by a factor of 2
b. Increases by a factor 4
c. Increases by a factor of 8
d. Stays the same; increase in v offset by decrease in time.
Since the initial kinetic energy is zero
E KE mv2
P


t
t
2t
Another darn ratio problem (FC#25)
The power needed to accelerate a projectile
from rest to its launch speed v in a time t
is P. By what factor does P change if the
same projectile is accelerated from rest to
a launch speed of 2v in a time of ½ t?
Assume no change in height.
a. Increases by a factor of 2
b. Increases by a factor 4
c. Increases by a factor of 8
d. Stays the same; increase in v offset by
decrease in time.
The ratio of power in case 2 to that in case
one is:
m ( 2v ) 2
P2 
2t
2
mv 2
P1 
2t