Physics 11
HW #6 Solutions
Chapter 6:
Focus On Concepts: 2, 11, 21, 22
Problems: 8, 24, 41, 43, 51, 54, 66, 80, 85
Focus On Concepts 6-2
(b) Work is positive when the force has a component in the direction of the
displacement. The force shown has a component along the −x and along the +y axis.
Therefore, displacements in these two directions involve positive work.
Focus On Concepts 6-11
(c) The gravitational force is a conservative force, and a conservative force does no net
work on an object moving around a closed path.
Focus On Concepts 6-21
(b) The principle of conservation of mechanical energy applies only if the work done by
the net nonconservative force is zero, Wnc = 0 J. When only a single nonconservative
force is present, it is the net nonconservative force, and a force that is perpendicular to the
displacement does no work.
Focus On Concepts 6-22
(a) A single nonconservative force is the net nonconservative force, and the work it does
is given by Wnc = ∆KE + ∆PE. Since the velocity is constant, ∆KE = 0 J. Therefore,
Wnc = ∆PE = mg(hf − h0). But hf − h0 = −325 m, since hf is smaller than h0. Thus,
Wnc = (92.0 kg)(9.80 m/s2)(−325 m).
Problem 6-8
REASONING AND SOLUTION According to Equation 6.1, W = Fs cos θ, the work is
a.
W = (94.0 N)(35.0 m) cos 25.0° = 2980 J
b.
W = (94.0 N)(35.0 m) cos 0° = 3290 J
Problem 6-24
REASONING The initial speed v0 of the skier can be obtained by applying the work-
=
W
energy theorem:
1 mv 2
f
2
− 12 mv02 (Equation 6.3). This theorem indicates that the initial
kinetic energy 12 mv02 of the skier is related to the skier’s final kinetic energy
work W done on the skier by the kinetic frictional force according to
1 mv 2
f
2
and the
1=
mv02
2
1 mv 2
f
2
−W
Solving for the skier’s initial speed gives
=
v0
vf2 −
2W
m
(1)
The work done by the kinetic frictional force is given by W = ( f k cosθ )s (Equation 6.1),
where fk is the magnitude of the kinetic frictional force and s is the magnitude of the skier’s
displacement. Because the kinetic frictional force points opposite to the displacement of the
skier, θ = 180°. According to Equation 4.8, the kinetic frictional force has a magnitude of
f k = µk FN , where µk is the coefficient of kinetic friction and FN is the magnitude of the
normal force. Thus, the work can be expressed as
=
W (=
f k cosθ )s ( µk FN cos180°)s
Substituting this expression for W into Equation (1), we have that
v0 = vf2 −
2 m F ( cos180° ) s
2W
= vf2 − k N
m
m
(2)
Since the skier is sliding on level ground, the magnitude of the normal force is equal to the
weight mg of the skier (see Example 10 in Chapter 4), so FN = mg. Substituting this relation
into Equation (2) gives
v0
=
vf2 −
2 mk m g ( cos180° ) s
=
m
vf2 − 2 mk g ( cos180° ) s
SOLUTION Since the skier comes to a halt, vf = 0 m/s. Therefore, the initial speed is
=
v0
)s
vf2 − 2 mk g ( cos180°=
) 4.5 m/s
( 0 m/s )2 − 2 ( 0.050 ) ( 9.80 m/s 2 ) ( cos180° ) ( 21 m
=
Problem 6-41
REASONING Since air resistance is being neglected, the only force that acts on the golf
ball is the conservative gravitational force (its weight). Since the maximum height of the
trajectory and the initial speed of the ball are known, the conservation of mechanical
energy can be used to find the kinetic energy of the ball at the top of the highest point.
The conservation of mechanical energy can also be used to find the speed of the ball
when it is 8.0 m below its highest point.
SOLUTION
a.
The conservation of mechanical energy, Equation 6.9b, states that
1 mv 2 + mgh = 1 mv 2 + mgh
f
f
0
0
2
2

KEf
Solving this equation for the final kinetic energy, KEf, yields
KE f =12 mv02 + mg ( h0 − hf
=
b.
1
2
)
( 0.0470 kg )( 52.0 m / s )2 + ( 0.0470 kg ) ( 9.80 m / s2 =
) ( 0 m − 24.6 m )
52.2 J
The conservation of mechanical energy, Equation 6.9b, states that
1 mv 2 + mgh
f
2 

f
Ef
=
1 mv 2 + mgh
0
2 

0
Ef
The mass m can be eliminated algebraically from this equation, since it appears as a
factor in every term. Solving for vf and noting that the final height is hf = 24.6 m – 8.0 m
= 16.6 m, we have that
vf =
v0 + 2 g ( h0 − hf ) =
2
( 52.0 m / s )2 + 2 ( 9.80 m / s2 ) ( 0 m − 16.6 m ) =
48.8 m / s
Problem 6-43
REASONING To find the maximum height H above the end of the track we will analyze
the projectile motion of the skateboarder after she leaves the track. For this analysis we will
use the principle of conservation of mechanical energy, which applies because friction and
air resistance are being ignored. In applying this principle to the projectile motion,
however, we will need to know the speed of the skateboarder when she leaves the track.
Therefore, we will begin by determining this speed, also using the conservation principle in
the process. Our approach, then, uses the conservation principle twice.
SOLUTION Applying the conservation of mechanical energy in the form of Equation 6.9b,
we have
We designate the flat portion of the track as having a height h0 = 0 m and note from the
drawing that its end is at a height of hf = 0.40 m above the ground. Solving for the final
speed at the end of the track gives
vf =
v02 + 2 g ( h0 − hf ) =
( 5.4 m/s )
2
+ 2 ( 9.80 m/s 2 ) ( 0 m ) − ( 0.40 m )  = 4.6 m/s
This speed now becomes the initial speed v0 = 4.6 m/s for the next application of the
conservation principle. At the maximum height of her trajectory she is traveling
horizontally with a speed vf that equals the horizontal component of her launch velocity.
Thus, for the next application of the conservation principle vf = (4.6 m/s) cos 48º. Applying
the conservation of mechanical energy again, we have
Recognizing that h0 = 0.40 m and hf = 0.40 m + H and solving for H give
1
2
mvf2 + mg ( 0.40 m ) + H =
v02 − vf2
=
H =
2g
( 4.6 m/s )
1
2
mv02 + mg ( 0.40 m )
− ( 4.6 m/s ) cos 48°
= 0.60 m
2 ( 9.80 m/s 2 )
2
2
Problem 6-51
REASONING
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.
b. When air resistance, a nonconservative 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.
SOLUTION
a. The conservation of mechanical energy, as expressed by Equation 6.9b, states that
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
( v02 − vf2 ) + h
0
g
Setting h0 = 0 m and vf = 0 m/s, the final height, in the absence of air resistance, is
vo2 − vf2
=
hf
=
2g
b. The work-energy theorem is
(18.0 m / s )2 − ( 0 m/s )2
=
2 9.80 m / s 2
(
)
16.5 m
Wnc =
(
1 mv 2
f
2
− 12 mv02
) + ( mgh
− mgh0 )
f
(6.6)
where Wnc is the nonconservative work done by air resistance. According to Equation 6.1,
the work can be written as Wnc = ( FR cos 180° ) s , where FR 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 θ = 180° and cos θ = –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
( vf2 − vo2 ) + mg ( hf − h0 )
Solving for FR gives
FR =
1m
2
( vf2 − vo2 ) + mg ( hf − h0 )
−s
( 0.750 kg ) ( 0 m/s )2 − (18.0 m/s )2  + ( 0.750 kg ) ( 9.80 m/s2 ) (11.8 m )
=
− (11.8 m )
1
2
2.9 N
Problem 6-54
REASONING We will use the work-energy theorem Wnc
= Ef − E0 (Equation 6.8) to
find the speed of the student. Wnc is the work done by the kinetic frictional force and is
negative because the force is directed opposite to the displacement of the student.
SOLUTION The work-energy theorem states that
(
)
Wnc = 12 mvf2 + mghf − 12 mv02 + mgh0
((((

((((((

Ef
E0
Solving for the final speed gives
vf
=
=
2Wnc
m
(6.8)
+ v0 − 2 g ( hf − h0 )
2
(
3
2 −6.50 × 10 J
83.0 kg
) + ( 0 m /s )
2
(
)
− 2 9.80 m /s 2 ( −11.8
=
m)
8.6 m / s
Problem 6-66
REASONING The average power is defined as the work divided by the time,
Equation 6.10a, so both the work and time must be known. The time is given. The work
can be obtained with the aid of the work-energy theorem as formulated in
Wnc =
(
1
2
)
mvf2 − 12 mv02 + ( mghf − mgh0 ) (Equation 6.6). Wnc is the work done by the lifting
force acting on the helicopter. In using this equation, we note that two types of energy are
changing: the kinetic energy
(
1
mv2
2
) and the gravitational potential energy (mgh).
The
kinetic energy is increasing, because the speed of the helicopter is increasing. The
gravitational potential energy is increasing, because the height of the helicopter is
increasing.
SOLUTION The average power is
P=
Wnc
t
(6.10a)
where Wnc is the work done by the nonconservative lifting force and t is the time. The work is
related to the helicopter’s kinetic and potential energies by Equation 6.6:
Wnc =
(
1
2
)
mvf2 − 12 mv02 + ( mghf − mgh0 )
Thus, the average power is
Wnc
=
P =
t
=
P
1
2
(
1 mv 2
f
2
)
− 12 mv02 + ( mghf − mgh0 )
=
t
1m
2
(v
2
f
)
− v02 + mg ( hf − h0 )
t
(810 kg ) ( 7.0 m /s )2 − ( 0 m /s )2  + (810 kg ) ( 9.80 m /s2 ) [8.2 m − 0 m ]
=
3.5 s
2.4 × 104 W
Problem 6-80
REASONING The change in gravitational potential energy for both the adult and the child
is ∆PE = mghf - mgh0, where we have used Equation 6.5. Therefore, ∆PE = mg(hf - h0). In
this expression hf - h0 is the vertical height of the second floor above the first floor, and its
value is not given. However, we know that it is the same for both staircases, a fact that will
play the central role in our solution.
SOLUTION Solving ∆PE = mg(hf - h0) for hf - h0, we obtain
=
hf − h0
( ∆PE )Adult
=
and hf − h0
mAdult g
( ∆PE )Child
mChild g
Since hf - h0 is the same for the adult and the child, we have
( ∆PE )Adult ( ∆PE )Child
=
mAdult g
mChild g
Solving this result for (∆PE)Child gives
( ∆PE )Adult mChild ( 2.00 ×103 J ) (18.0 kg )
=
( ∆PE
)Child
=
mAdult
81.0 kg
=
444 J
Problem 6-85
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.
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, hf = h , and h0 = H .
Since the swimmer starts from rest, v0 = 0 m/s, and the above expression becomes
1 2
v
2 f
+ gh = gH
Solving for H, we obtain
v2
H = h+ f
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
The vertical displacement of the swimmer after leaving the slide is, from Equation 3.5b
(with down being negative),
2
2
2
y = 2 ay t = 2 (–9.80 m/s )(0.500 s) = −1.23 m
1
1
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
= 1.23 m +
= 6.33 m
H = h+
2g
2(9.80 m/s2 )