Ch6. Work and Energy
Work Done by a Constant Force
Force F points in the same direction as the
resulting displacement s
W=Fs
1
W  ( F cos )s
SI Unit of Work: newton.meter=joule(J)
Units of Measurement for Work
System
Force
SI
newton(N) meter(m)
CGS
dyne(dyn)
centimeter(cm)
erg
BE
pound(lb)
foot(ft)
foot.pound
*
Distance =
Work
joule(J)
(ft.lb)
2
Example 1. Pulling a Suitcaseon-Wheels
3
Find the work done by a 45.0 N force in
pulling the suitcase at an angle   50.0 for
a distance s=75.0 m
W  ( F cos )s  (45.0 N ) cos50.0(75.0m)  2170J
No work done due to Fsin 
F and s in the same direction
positive work
F and s in the opposite direction
negative work
4
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.
5
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) W  ( F cos )s  (710N ) cos0(0.65m)  460J
(b) W  ( F cos )s  (710N ) cos180(0.65m)  460J
cos1800 = -1
6
Example 3. Accelerating a Crate
A 120kg crate on the flatbed of a
truck that is moving with an
acceleration of a=+1.5m/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
s=65m. What is the total work
done on the crate by all of the
forces acting on it?
7
Forces that act on the crate:
(1) the weight W=mg of the crate,
(2) the normal force FN exerted by the flatbed,
(3) the static frictional force fs.
f s  ma  (120kg)(1.5m / s )  180N
2
W  ( f s cos )s  (180N ) cos0(65m)  1.2 10 J
4
8
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?
(a) The work is negative when you ride up the escalator and
positive when you ride down the escalator.
(b) The work is positive when you ride up the escalator and
negative when you ride down the escalator.
(c) The work is positive irrespective of whether you ride up or
down the escalator.
(d) The work is negative irrespective of whether you ride up
or down the escalator.
Answer: (b)
9
The Work-Energy Theorem and
Kinetic Energy
10
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.
11
v  v  2as  s 
2
f
2
0
v v
2
f
2
0
2a
Definition of Kinetic Energy:
The kinetic energy KE of an object with mass m
and speed v is given by
1 2
KE  mv
2
SI Unit of Kinetic Energy: joule(J)
12
( F)s  mas
Work done by net ext. force
( F ) s
Work done by
net ext. force
1 2 1 2
 mv f  mv 0
2
2
Final KE
Initial KE
13
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:
1 2 1 2
W  KE f  KE0  mv f  mv 0
2
2
14
Example 4. Deep Space 1
15
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 v0=275m/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 2.42*109 m. Determine the final speed of the
probe, assuming that the mass remains nearly constant.
16
W  ( F cos )s  (56.0 103 N ) cos0(2.42109 m)
=1.36*108 J
1
KE f  W  KE0  (1.36 10 J )  (474 kg )( 275 m / s) 2
2
8
=1.54*108 J
vf 
2( KE f )
m
2(1.54108 J )

=806 m/s
474kg
17
Example 5. Downhill Skiing
18
A 58 kg skier is coasting down a 250 slope. A kinetic
frictional force of magnitude fk=70N opposes her
motion. Near the top of the slope, the skier’s speed
is v0=3.6m/s. Ignoring air resistance, determine the
speed vf at a point that is displaced 57m downhill.
2
F

mg
sin
25


f

(
58
kg
)(
9
.
80
m
/
s
) sin 25  70N

k
=+170 N
19
W  ( F cos )s  (170N ) cos0(57m)  9700J
KEf=W+KE0=9700J+(1/2)(58kg)(3.6m/s)2=10100J
vf 
2( KE f )
m
2(10100J )

 19m / s
58kg
20
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 rocket’s engine.”
Answer: False
21
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.
In circular motion, F
no work done.
S always
KE changes in the elliptical orbit,
but not in the circular orbit.
22
Gravitational Potential Energy
Work Done by the Force of Gravity
Wgravity=(mg cos00)(h0-hf)=mg(h0-hf)
23
Example 7. A Gymnast on a
Trampoline
24
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.
Wgravity=mg (h0-hf)
v0   2 g (h0  h f )   2(9.80m / s 2 (1.20m  4.80m)
=8.40 m/s
25
Gravitational Potential Energy
Wgravity= mgh0 Initial
gravitationa
l potential
energy PE0
mghf
Final
gravitational
potential
energy PEf
26
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:
PE=mgh
SI Unit of gravitational potential energy: joule (J)
27
Conservative Versus Nonconservative Forces
28
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 object’s 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.
29
Work done by external forces =
i.e,
KE
1 2 1 2
Wc  Wnc  mv f  mv 0
2
2
1 2 1 2
mg (h0  h f )  Wnc  mv f  mv 0
2
2
30
1 2 1 2
Wnc  ( mv f  mv 0 )  (mgh f  mgh 0 )
2
2
Wnc
=
Net work
done by nonconservative
forces
(KEf - KE0)
Change in
kinetic energy
+
(PEf - PE0)
Change in gravitational
potential energy
Wnc  KE  PE
31
The Conservation of Mechanical
Energy
32
Total mechanical energy: the sum of kinetic
energy and gravitational potential energy
E = KE + PE
Wnc
Wnc
=
(KEf - KE0)
=
+
(PEf - PE0)
(KEf + PEf) - (KE0 + PE0)
Ef
E0
Wnc = Ef - E0
33
Suppose Wnc=0J, so Ef = E0
1
2
mvf2+mghf =
1
2
mv02+mgh0
The principle of Conservation of Mechanical Energy
The total mechanical energy (E=KE+PE) of an
object remains constant as the object moves,
provided that the net work done by external nonconservative forces is zero, Wnc=0J
34
35
36
Example 8. A Daredevil
Motorcyclist
37
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)mvf2+mghf = (1/2) mv02+mgh0
v f  v02  2 g (h0  h f )
v f  (38.9m / s)  2(9.80 m / s )( 70.0m  35.0m)
2
2
=46.2 m/s
38
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?
(a) An object moves uphill with an increasing speed.
(b) An object moves uphill with a decreasing speed.
(c) An object moves uphill with a constant speed.
(d) An object moves downhill with an increasing speed.
(e) An object moves downhill with a decreasing speed.
(f) An object moves downhill with a constant speed.
(b) And (d)
39
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.
40
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?
If Wnc= 0 (work due to nonconservative forces)
conservation of energy
Tension and air resistance --- non conservative
T is always
r to the circular path.
So no work done by T.
41
Work done by air resistance is nonzero.
So ideally no.
But if ignore air resistance,
1 2
1 2
mv 0  mgh 0  mv f  mgh f
2
2
v f   v  2 g (h0  h f )
2
0
42
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)mvf2+mghf = (1/2) mv02+mgh0
Ef
E0
43
v f  v0  2 g (h0  h f )
2
vf 
(3.0m / s)  2(9.80 m / s )( 93.5m)
2
= 42.9m/s (about 96 mi/h)
44
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
45
1
1
2
2
Wnc  ( mv f  mgh f )  ( mv 0  mgh 0 )
2
2
Ef
E0
1
2
2
Wnc  m(v f  v0 )  mg (h0  h f )
2
1
Wnc  (55.0kg )[( 41m / s ) 2  (3m / s ) 2 ]  (55.0kg )( 9.80 m / s 2 )( 93.5m)
2
 4400 J
46
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.
47
1
1
2
2
Wnc  ( mv f  mgh f )  ( mv 0  mgh 0 )
2
2
1
2
2[Wnc  m v0  m g(h f  h0 )]
2
vf 
m
1
2[425J  (0.20kg )(0m / s) 2  (0.20kg )(9.80m / s 2 )(29m)]
2
vf 
0.20kg
=61m/s
48
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.
Work W
P

Time
t
SI Unit of Power: joule/s=watt (W)
49
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?
(a) Engine A and B can do the same amount of work in the
same amount of time.
(b) In the same amount of time, engine B can do more work
than engine A.
(c) Engine A and B can do the same amount of work, but A can
do it more quickly.
Answer: (c)
50
Units of Measurement for Power
System
Work / Time
SI
joule (J) / second (s) =
watt (W)
CGS
erg / second (s)
BE
foot.pound / second (s) =
erg per
second(erg/s)
foot.pound
per second
(ft.lb)
=
=
Power
(ft.lb/s)
51
P
change
in energy
Time
1horsepower 550 foot  pound/ sec ond  745.7watts
W Fs

t
t
P  Fv
52
Example 13. The Power to
Accelerate a Car
A 1.10*103kg car, starting from rest, accelerates for
5.00s. The magnitude of the acceleration is
a=4.60m/s2. Determine the average power generated
by the net force that accelerates the vehicle.
F  ma  (1.1010 kg)(4.60m / s )  5060N
3
2
53
1
1
v  (v 0  v f )  v f
2
2
v f  v0  at  (0m / s)  (4.60m / s 2 )(5.00s)  23.0m / s
P  F v  (5060N )(11.5m / s)  5.82104W (78.0hp)
54
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.
55
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.
56
W  ( F cos )1 s1  ( F cos ) 2 s2    
57
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
J
W  (242squares)(0.250
)  60.5 J
square
58
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.
59
60
W= (Fcos) s
Force
F
Angle
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)+0J=+1470J
61
Concepts & Calculations Example 16.
Conservation of Mechanical Energy and
the Work-Energy Theorem
hA
62
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.
(a) What is the kinetic energy of the block when it
reaches B?
(b) How much work does the kinetic frictional force do
during the BC segment of the trip?
63
A---B motion: What are the forces?
Weight--------------- conservative
Normal force-------- non conservative
What is the non conservative work (Wnc) ?
Wnc=0
FN
displacement
Conservation of energy valid for A---B?
K. E at B
?
K. E at A
K. E at B
<
K. E at A
64
(a) KEB + mghB = KEA + mghA
KEB = KEA + mg(hA-hB)
=37J+(0.41kg)(9.80m/s2)(12m-7m)
=57J
65
(b)
For BC Trip, is total energy conserved? Why?
No
Then
(Friction
Wnc)
Wnc= KEC + mghC - (KEB+mghB)
Wnc= KEC - KEB+mg(hC-hB)
0J
0m
Wnc= -KEB= -57J
66
Problem 2
T
T
REASONING AND SOLUTION Each locomotive does work
W = T(cos) s= (5.00*103 N) cos 20.0° (2.00 * 103 m) = 9.40*106 J
The net work is then
WT = 2W = 2T(cos) s = 1.88 × 107 J
67
Problem 16
8450 m/s
perigee
2820 m/s
apogee
68
From the work-energy theorem,
W
1
2
mvf
2

1
2
mv0
2

1  2
m vf
2 
2
 v0

a)
11
11
b) W  1 (7420 kg) (2820 m/s)2  (8450 m/s)2  = –2.35
2
.
35
10
10
JJ
2


69
Problem 20
16 kg
F = 24N
8.0 M
v0 = 0
vf = 2.0 m/8
FN
P
fk
fk = μk FN
mg
FN-mg=0
FN=mg
To find the work, employ the work-energy theorem,
W = KEf  KE0
70
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 θ) s] to express each work
2
1 , and
contribution, writing the kinetic energy as
mv
2
noting that the initial kinetic energy is zero (the sled starts
from rest), we obtain
 P cos 0 ) s   f k cos180 ) s 
Wpull
Wf
1
2
mv
2
71
The angle θ 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
 P cos 0) s   k mg cos180) s 
Wpull
1
2
mv
2
Wf
Solving for the coefficient of kinetic friction gives
mv   P cos 0 ) s
k 

 mg cos180) s
1
2
2
1
2
16 kg ) 2.0 m/s )   24 N )8.0 m ) 
 16 kg )  9.80 m/s 2 )  8.0 m )
2
72
0.13
Problem 26
REASONING The work done by the weight of
the basketball is given by Equation 6.1
as W   F cos  ) s , where F = mg is the
magnitude of the weight,  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  = 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
73
SOLUTION
a. The work done by the weight of the basketball is
W   F cos  ) s = mg (cos 0)(h0  hf)
= (0.60 kg)(9.80 m/s2)(6.1 m  1.5 m) =
27 J
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) = 36 J
74
c. The potential energy of the ball, relative to the ground,
when it is caught is
PEf = mghf = (0.60 kg)(9.80 m/s2)(1.5 m) = 8.8 J
d. The change in the ball’s gravitational potential energy is
PE = PEf  PE0 = 8.8 J – 36 J = 27 J
We see that the change in the gravitational potential
energy is equal to –27 J = W , where W is the work
done by the weight of the ball (see part a).
75
Problem 34
Total mechanical energy
conserved?
Yes (W—conservative
FN
1 mv 2
f
2
hf  h0
14.0 m/s )



2
 mghf 
 13.0 m/s )
2 9.80 m/s
2
)
displacement)
1 mv 2
0
2
 mgh0
2
 1.4 m
76
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 nonconservative forces, and the
principle of conservation of
mechanical energy applies.
77
SOLUTION Applying the principle of conservation of
mechanical energy to the swimmer at the top and the bottom
of the slide, we have
1
1
2
2
mvf  mghf  mv0  mgh0
2
2
Ef
E0
If we let h be the height of the bottom of the slide above the
water, hf  h, and h0  H. Since the swimmer starts from
rest, v0  0 m/s, and the above expression becomes
1 2
vf
2
 gh  gH
78
Solving for H, we obtain
vf2
H  h
2g
Before we can calculate H, we must find vf and h.
Since the velocity in the horizontal direction is
constant,
x 5.00 m
vf 

10.0 m /s
t 0.500 s
79
The vertical displacement of the swimmer after
leaving the slide is, from Equation 3.5b (with down
being negative),

)
1 2 1
2
2
y  a y t   9.80 m / s 0.500 s )  1.23m
2
2
Therefore, h = 1.23 m. Using these values of vf and h
in the above expression for H, we find
vf2
(10.0 m/s)2
H  h
 1.23 m 
2  6.33 m
2g
2(9.80 m/s )
80
Problem 43
60
r
rcos60
r
PE=0
mv
r
FC
2
 T  mg
m v2
T  mg
r
This is the max tension.
81
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,
mv
r
2
 T  mg
FC
Solving for the mass yields
m
T
2
v
g
r
82
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
vf = 2gh0  gr
1r
h0 = r (1 – cos 60.0°) = 2
r=2h0
83
The expression for the mass now becomes
T
T
m 2

2 gh0
v
g
g
r
r
T
T


g  g 2g
2
8.00  10 N
T
m

 40.8 kg
2 g 2 9.80 m/s2

)
84
Problem 48
No air friction.
18.0 m/s
m=0.75kg
85
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.
The conservation of mechanical energy, as expressed by
Equation 6.9b, states that
1 mv 2
f
2
 mghf 
Ef
1 mv 2
0
2
 mgh0
E0
86
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
hf 
1
2

2
v0

2
vf
)h
0
g
Setting h0 = 0 m and vf = 0 m/s, the final height, in the
absence of air resistance, is
hf 
2
vo

2g
2
vf
18.0 m / s )   0 m/s )
2


2 9.80 m / s
2
)
2
 16.5 m
87
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.
The work-energy theorem is
Wnc 

1 mv 2
f
2

1 mv 2
0
2
)   mgh
f
 mgh0 )
88
where Wnc is the non-conservative work done by air
resistance. According to Equation 6.1, the work can be
written as Wnc  (F R cos180)s , where F is the average
R
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  = 180 and cos  = –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
 FR s 
1m
2

2
vf
2
 vo
)  mg  hf  h0 )
89
Solving for FR gives
FR 


1 m v2
f
2
)
 vo2  mg  hf  h0 )
s
1
2

)
2
2
2

0.750
kg
0
m/s

18.0
m/s

0.750
kg
9.80
m/s

)  ) 
) 
)
11.8 m )
 11.8 m )
 2.9 N
90
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 W  P t . Using the fact
that 1 kW = 1.0 10 3 J/s and that 1h = 3600 s, we have
3
3
6
1.0 kWh = (1.0 10 J/s)(1 h) = (1.0 10 J/s)(3600 s)= 3.6 10 J
91
Problem 60
friction
REASONING AND SOLUTION
mg
37
a. The power developed by the engine is
P = Fv = (2.00 * 102 N)(20.0 m/s) = 4.00 103 W
92
b. The force required of the engine in order to
maintain a constant speed up the slope is
F = Fa + mg sin 37.0°
The power developed by the engine is then
P = Fv = (Fa + mg sin 37.0°)v
P = [2.00 * 102 N + (2.50 * 102 kg)(9.80 m/s2)sin 37.0°](20.0 m/s)
=
3.35 10 4 W
93