Ch6. Work and Energy Work Done by a Constant Force

1
Ch6. Work and Energy Work Done by a Constant
Force
Force F points in the same direction as the
resulting displacement s
WFs
2
SI Unit of Work newton.meterjoule(J)
Units of Measurement for Work
System Force Distance Work
SI newton(N) meter(m) joule(J)
CGS dyne(dyn) centimeter(cm) erg
BE pound(lb) foot(ft) foot.pound (ft.lb)
3
Example 1. Pulling a Suitcase-on-Wheels
4
Find the work done by a 45.0 N force in pulling
the suitcase at an angle for a
distance s75.0 m
No work done due to Fsin
F and s in the same direction
positive work F and s in the opposite direction
negative work
5
Example 2. Bench-Pressing
The weight lifter is bench-pressing a barbell
whose weight is 710N. In part (b) of the figure,
he raises the barbell a distance of 0.65m above
his chest, and in part (c) he lowers it the same
distance.
6
The weight is raised and lowered at a constant
velocity. Determine the work done on the barbell
by the weight lifter during (a) the lifting phase
and (b) the lowering phase.
(a)
(b)
cos1800 -1
7
Example 3. Accelerating a Crate
A 120kg crate on the flatbed of a truck that is
moving with an acceleration of a1.5m/s2 along
the positive x axis. The crate does not slip with
respect to the truck, as the truck undergoes a
displacement whose magnitude is s65m. What is
the total work done on the crate by all of the
forces acting on it?
8

• Forces that act on the crate
• the weight Wmg of the crate,
• the normal force FN exerted by the flatbed,
• the static frictional force fs.

9
Check your understanding 1

• A suitcase is hanging straight down from your
  hand as you ride an escalator. Your hand exerts a
  force on the suitcase and this force does
  work.Which one of the following statements is
  correct?
• The work is negative when you ride up the
  escalator and positive when you ride down the
  escalator.
• The work is positive when you ride up the
  escalator and negative when you ride down the
  escalator.
• The work is positive irrespective of whether you
  ride up or down the escalator.
• The work is negative irrespective of whether you
  ride up or down the escalator.

Answer (b)
10
The Work-Energy Theorem and Kinetic Energy
11
Concepts At a Glance In physics, when a net
force performs work on an object, there is always
a result from the effort. The result is a change
in the Kinetic energy. What is Kinetic
Energy? Energy associated with motion.
KE(1/2)mv2 The relationship that
relates work to the change in kinetic energy is
known as the Work-energy theorem.
12
Definition of Kinetic Energy The kinetic energy
KE of an object with mass m and speed v is given
by
SI Unit of Kinetic Energy joule(J)
13
Work done by net ext. force
14
The Work-Energy Theorem When a net external force
does work W on an object, the kinetic energy of
the object changes from its initial value of KE0
to a final value of KEf, the difference between
the two values being equal to the work
15
Example 4. Deep Space 1
16
The space probe Deep Space 1 was launched October
24, 1998. Its mass was 474kg. The goal of the
mission was to test a new kind of engine called
an ion propulsion drive, which generates only a
weak thrust, but can do so for long periods of
time using only small amounts of fuel. The
mission has been spectacularly successful.
Consider the probe traveling at an initial speed
of v0275m/s. No forces act on it except the
56.0-mN thrust of its engine. This external force
F is directed parallel to the displacement s of
magnitude 2.42109 m. Determine the final speed
of the probe, assuming that the mass remains
nearly constant.
17
1.36108 J
1.54108 J
806 m/s
18
Example 5. Downhill Skiing
19
A 58 kg skier is coasting down a 250 slope. A
kinetic frictional force of magnitude fk70N
opposes her motion. Near the top of the slope,
the skiers speed is v03.6m/s. Ignoring air
resistance, determine the speed vf at a point
that is displaced 57m downhill.
170 N
20
KEfWKE09700J(1/2)(58kg)(3.6m/s)210100J
21
Check your understanding 2
A rocket is at rest on the launch pad. When the
rocket is launched, its kinetic energy increases.
Is the following statement true or false? The
amount by which the kinetic energy increases is
equal to the work done by the force generated by
the rockets engine.
Answer False
22
Conceptual Example 6. Work and Kinetic Energy
A satellite moving abut the earth in a circular
orbit and in an elliptical orbit. The only
external force that acts on the satellite is the
gravitational force. For these two orbits,
determine whether the kinetic energy of the
satellite changes during the motion.
KE changes in the elliptical orbit, but not in
the circular orbit.
23
Gravitational Potential Energy Work Done by the
Force of Gravity
Wgravity(mg cos00)(h0-hf)mg(h0-hf)
24
Example 7. A Gymnast on a Trampoline
25
A gymnast springs vertically upward from a
trampoline. The gymnast leaves the trampoline at
a height of 1.20m and reaches a maximum height of
4.80m before falling back down. All heights are
measured with respect to the ground. Ignoring air
resistance, determine the initial speed v0 with
which the gymnast leaves the trampoline.
Wgravitymg (h0-hf)
8.40 m/s
26
Gravitational Potential Energy
Wgravity mgh0 - mghf
Initial gravitational potential energy PE0
Final gravitational potential energy PEf
27
Definition of Gravitational Potential Energy
The gravitational potential energy PE is the
energy that an object of mass m has by virtue of
its position relative to the surface of the
earth. That position is measured by the height h
of the object relative to an arbitrary zero
level PEmgh SI
Unit of gravitational potential energy joule (J)
28
Conservative Versus Non-conservative Forces
29
Definition of a Conservative Force
Version 1 A force is conservative when the work
it does on a moving object is independent of the
path between the objects initial and final
positions.
Version 2 A force is conservative when it does
no net work on an object moving around a closed
path, starting and finishing at the same point.
30
(No Transcript)
31
(No Transcript)
32
The Conservation of Mechanical Energy
33
Total mechanical energy the sum of kinetic
energy and gravitational potential energy
E KE PE
Wnc (KEf - KE0) (PEf
- PE0)
Wnc (KEf PEf) - (KE0 PE0)
Ef
E0
Wnc Ef - E0
34
Suppose Wnc0J, so Ef E0
mvf2mghf mv02mgh0
The principle of Conservation of Mechanical Energy
The total mechanical energy (EKEPE) of an
object remains constant as the object moves,
provided that the net work done by external
non-conservative forces is zero, Wnc0J
35
(No Transcript)
36
(No Transcript)
37
Example 8. A Daredevil Motorcyclist
38
A motorcyclist is trying to leap across the
canyon by driving horizontally off the cliff at a
speed of 38.0 m/s. Ignoring air resistance, find
the speed with which the cycle strikes the ground
on the other side.
(1/2)mvf2mghf (1/2) mv02mgh0
46.2 m/s
39
Check your understanding 3

• Some of the following situations are consistent
  with the principle of conservation of mechanical
  energy, and some are not. Which ones are
  consistent with the principle?
• An object moves uphill with an increasing speed.
• An object moves uphill with a decreasing speed.
• An object moves uphill with a constant speed.
• An object moves downhill with an increasing
  speed.
• An object moves downhill with a decreasing speed.
• An object moves downhill with a constant speed.

(b) And (d)
40
Conceptual Example 9. The Favorite Swimming Hole
A rope is tied to a tree limb and used by a
swimmer to swing into the water below. The person
starts from rest with the rope held in the
horizontal position, swings downward, and then
lets go of the rope.
41
Three forces act on him his weight, the tension
in the rope, and the force due to air resistance.
His initial height h0 and final height hf are
known. Considering the nature of these forces,
conservative versus non-conservative, can we use
the principle of conservation of mechanical
energy to find his speed vf at the point where he
lets go of the rope?
Tension and air resistance --- non conservative
So no work done by T.
42
Work done by air resistance is nonzero.
So ideally no.
But if ignore air resistance,
43
Example 10. The Steel Dragon
The tallest and fastest roller coaster in the
world is now the Steel Dragon in Mie, Japan. The
ride includes a vertical drop of 93.5m. The
coaster has a speed of 3.0m/s at the top of the
drop. Neglect friction and find the speed of the
riders at the bottom.
(1/2)mvf2mghf (1/2) mv02mgh0
Ef
E0
44
42.9m/s (about 96 mi/h)
45
Example 11. The Steel Dragon, Revisited
In example 10, we ignored non-conservative
forces, such as friction. In reality, however,
such forces are present when the roller coaster
descends. The actual speed of the riders at the
bottom is 41.0 m/s, which is less than that
determined in example 10. Assuming again that the
coaster has a speed of 3.0 m/s at the top, find
the work done by non-conservative forces on a
55.0kg rider during the descent from a height h0
to a height hf , where h0 hf 93.5m
46
E0
Ef
47
Example 12. Fireworks
A 0.20kg rocket in a fireworks display is
launched from rest and follows an erratic flight
path to reach the point P. Point P is 29m above
the starting point. In the process, 425J of work
is done on the rocket by the non-conservative
force generated by the burning propellant.
Ignoring air resistance and the mass lost due to
the burning propellant, find the speed vf of the
rocket at the point P.
48
61m/s
49
Power
The idea of power incorporates both the concepts
of work and time, for power is work done per unit
time.
Definition of Average Power
Average power P is the average rate at which work
W is done, and it is obtained by dividing W by
the time t required to perform the work.
SI Unit of Power joule/swatt (W)
50
Check your understanding 4

• Engine A has a greater power rating than engine
  B. Which one of the following statements
  correctly describes the abilities of these
  engines to do work?
• Engine A and B can do the same amount of work in
  the same amount of time.
• In the same amount of time, engine B can do more
  work than engine A.
• Engine A and B can do the same amount of work,
  but A can do it more quickly.

Answer (c)
51
Units of Measurement for Power
System Work / Time Power
SI joule (J) / second (s) watt (W)
CGS erg / second (s) erg per second(erg/s)
BE foot.pound / second (s) foot.pound per second
(ft.lb) (ft.lb/s)
52
(No Transcript)
53
Example 13. The Power to Accelerate a Car
A 1.10103kg car, starting from rest, accelerates
for 5.00s. The magnitude of the acceleration is
a4.60m/s2. Determine the average power generated
by the net force that accelerates the vehicle.
54
(No Transcript)
55
Other Forms of Energy and the Conservation of
Energy
Examples electrical energy, heat, chemical
energy, nuclear energy.
In general, energy of all types can be converted
from one form to another.
The principle of conservation of energy Energy
can neither be created nor destroyed, but can
only be converted from one form to another.
56
Work Done by a Variable Force
The work done by a variable force in moving an
object is equal to the area under the graph of
Fcos versus s.
57
(No Transcript)
58
Example 14. Work and the Compound Bow
Find the work that the archer must do in drawing
back the string of the compound bow from 0 to
0.500 m
59
Example 15. Skateboarding and Work
The skateboarder is coasting down a ramp, and
there are three forces acting on her her weight
W (magnitude 675 N), a frictional force f (
magnitude 125 N). That opposes her motion, and
a normal force FN (magnitude 612 N). Determine
the net work done by the three forces when she
coasts for a distance of 9.2 m.
60
(No Transcript)
61
Force F Angle S W (Fcosq) s
W 675N 65.00 9.2m W(675N)(cos 65.00)(9.2m)2620J
f 125N 180.00 9.2m W(125N)(cos 180.00)(9.2m)-1150J
FN 612N 90.00 9.2m W(612N)(cos 90.00)(9.2m)0J
The net work done by the three forces is
2626J(-1150J)0J1470J
62
Concepts Calculations Example 16. Conservation
of Mechanical Energy and the Work-Energy Theorem
hA
63

• A 0.41kg block sliding from A to B along a
  frictionless surface. When the block reaches B,
  it continues to slide along the horizontal
  surface BC. The block slows down, coming to rest
  at C. The kinetic energy of the block at A is
  37J, and the height of A and B are 12m and 7m
  above the ground.
• What is the kinetic energy of the block when it
  reaches B?
• How much work does the kinetic frictional force
  do during the BC segment of the trip?

64
A---B motion What are the forces?
Weight---------------
conservative
Normal force--------
non conservative
What is the non conservative work (Wnc) ?
Conservation of energy valid for A---B?
K. E at B ? K. E at A
K. E at B lt K. E at A
65
(a) KEB mghB KEA mghA
KEB KEA mg(hA-hB) 37J(0.41kg)(9.80m
/s2)(12m-7m) 57J
66
(b) For BC Trip, is total energy conserved?
Why?
67
Problem 2
T
T
The net work is then WT 2W
2T(cosq) s  1.88  107 J 
68
Problem 16
2820 m/s apogee
8450 m/s perigee
69
From the work-energy theorem,
a)
b)
70
Problem 20
F 24N
16 kg
8.0 M
v0 0
vf 2.0 m/8
FN
fk µk FN
FN-mg0 FNmg
P
fk
mg
To find the work, employ the work-energy theorem,
W KEf ?
KE0
71
Work done by the net force (pulling force P and
fk) W W pulling W f
SOLUTION According to the work-energy theorem,
we have W Wpull Wf
KEf ? KE0
Using Equation 6.1 W  (F cos ?) s to express
each work contribution, writing the kinetic
energy as , and noting that the initial
kinetic energy is zero (the sled starts from
rest), we obtain
72
The angle ? between the force and the
displacement is 0º for the pulling force (it
points in the same direction as the displacement)
and 180º for the frictional force (it points
opposite to the displacement). Equation 4.8
indicates that the magnitude of the frictional
force is fk µkFN, and we know that the
magnitude of the normal force is FN mg. With
these substitutions the work-energy theorem
becomes
Solving for the coefficient of kinetic friction
gives
73
Problem 26
REASONING The work done by the weight of the
basketball is given by Equation 6.1 as
, where F mg is the magnitude
of the weight, q is the angle between the weight
and the displacement, and s is the magnitude of
the displacement. The drawing shows that the
weight and displacement are parallel, so that q
0?. The potential energy of the basketball is
given by Equation 6.5 as PE mgh, where h is the
height of the ball above the ground.
6.1 m
mg
s
0.6g
1.5 m
74
SOLUTION
a. The work done by the weight of the basketball
is
mg (cos 0?)(h0 ? hf) (0.60 kg)(9.80
m/s2)(6.1 m ? 1.5 m)
b. The potential energy of the ball, relative to
the ground, when it is released is
PE0 mgh0 (0.60 kg)(9.80 m/s2)(6.1 m)
75

• d. The change in the balls gravitational
  potential energy is
•   DPE PEf ? PE0 8.8 J 36 J ?27 J

We see that the change in the gravitational
potential energy is equal to 27 J
, where W is the work done by the weight of the
ball (see part a).
76
Problem 34
Total mechanical energy conserved?
77
Problem 41
REASONING Friction and air resistance are being
ignored. The normal force from the slide is
perpendicular to the motion, so it does no work.
Thus, no net work is done by non-conservative
forces, and the principle of conservation of
mechanical energy applies.
78
SOLUTION Applying the principle of
conservation of mechanical energy to the swimmer
at the top and the bottom of the slide, we have
If we let h be the height of the bottom of the
slide above the water, , and
. Since the swimmer starts from rest,
m/s, and the above expression becomes
79
Solving for H, we obtain
Before we can calculate H, we must find and
h. Since the velocity in the horizontal
direction is constant,
80
The vertical displacement of the swimmer after
leaving the slide is, from Equation 3.5b (with
down being negative),
Therefore, h 1.23 m. Using these values of
and h in the above expression for H, we find
81
Problem 43
r
60
rcos60
r
PE0
This is the max tension.
82
REASONING AND SOLUTION At the bottom of the
circular path of the swing, the centripetal force
is provided by the tension in the rope less the
weight of the swing and rider. That is,
Solving for the mass yields
83
The energy of the swing is conserved if friction
is ignored. The initial energy, E0, when the
swing is released is completely potential energy
and is E0 mgh0, Conservation of energy
PEini KEini PEf KEf
mgh0 0 0 (1/2)mvf2
r2h0
84
The expression for the mass now becomes
85
Problem 48
No air friction.
18.0 m/s
m0.75kg
86
a. Since there is no air friction, the only force
that acts on the projectile is the conservative
gravitational force (its weight). The initial and
final speeds of the ball are known, so the
conservation of mechanical energy can be used to
find the maximum height that the projectile
attains.
87
The mass m can be eliminated algebraically from
this equation since it appears as a factor in
every term. Solving for the final height hf gives
Setting h0 0 m and vf 0 m/s, the final
height, in the absence of air resistance, is
88
b. When air resistance, a non-conservative force,
is present, it does negative work on the
projectile and slows it down. Consequently, the
projectile does not rise as high as when there is
no air resistance. The work-energy theorem, in
the form of Equation 6.6, may be used to find the
work done by air friction. Then, using the
definition of work, Equation 6.1, the average
force due to air resistance can be found.
89
where Wnc is the non-conservative work done by
air resistance. According to Equation 6.1, the
work can be written as
, where is the average force of air
resistance. As the projectile moves upward, the
force of air resistance is directed downward, so
the angle between the two vectors is q 180? and
cos q 1. The magnitude s of the displacement
is the difference between the final and initial
heights, s hf h0 11.8 m. With these
substitutions, the work-energy theorem becomes
90
Solving for gives
91
Problem 57
REASONING AND SOLUTION One is the amount of
work or energy generated when one kilowatt of
power is supplied for a time of one hour. From
Equation 6.10a, we know that .
Using the fact that
and that 1h  3600 s, we have
92
Problem 60
friction
REASONING AND SOLUTION
mg
37

1. The power developed by the engine is

93
b. The force required of the engine in order to
maintain a constant speed up the slope is
F Fa mg sin 37.0