Week 5 homework
IMPORTANT NOTE ABOUT WEBASSIGN:
In the WebAssign versions of these problems, various details have been changed, so that
the answers will come out differently. The method to find the solution is the same, but
you will need to repeat part of the calculation to find out what your answer should have
been.
WebAssign Problem 1: During a tug-of-war, team A pulls on team B by applying a
force of 1100 N to the rope between them. How much work does team A do if they pull
team B toward them a distance of 2.0 m?
REASONING AND SOLUTION We will assume that the tug-of-war rope remains
parallel to the ground, so that the force that moves team B is in the same direction as the
displacement. According to Equation 6.1, the work done by team A is
W = (F cosθ )s = (1100 N)(cos 0°)(2.0 m) = 2.2 × 10 3 J
WebAssign Problem 2: A 55-kg box is being pushed a distance of 7.0 m across the floor
by a force whose magnitude is 150 N. The force is parallel to the displacement of the
box. The coefficient of kinetic friction is 0.25. Determine the work done on the box by
each of the four forces that act on the box. Be sure to include the proper plus or minus
sign for the work done by each force.
REASONING AND SOLUTION The applied force does work
WP = Ps cos 0° = (150 N)(7.0 m) = 1.0 × 103 J
The frictional force does work
Wf = fks cos 180° = – µ kFNs
where FN = mg, so
2
Wf = – (0.25)(55 kg)(9.80 m/s )(7.0 m) = –940 J
, since they both act at a 90° angle to the displacement.
WebAssign Problem 3: The concepts in this problem are similar to those in MultipleConcept Example 4, except that the force doing the work in this problem is the tension in
the cable. A rescue helicopter lifts a 79-kg person straight up by means of a cable. The
person has an upward acceleration of 0.70 m/s2 and is lifted from rest through a distance
of 11 m. (a) What is the tension in the cable? How much work is done by (b) the tension
in the cable and (c) the person’s weight? (d) Use the work–energy theorem and find the
final speed of the person.
REASONING Since the person has an upward acceleration, there must be a net force
acting in the upward direction. The net force ΣFy is related to the acceleration ay by
Newton’s second law, Σ Fy = ma y , where m is the mass of the person. This relation will
allow us to determine the tension in the cable. The work done by the tension and the
person’s weight can be found directly from the definition of work, Equation 6.1.
SOLUTION
a. The free-body diagram at the right shows the two forces that act on
the person. Applying Newton’s second law, we have
T
+y
s
T − mg = ma y
1
424
3
Σ Fy
Solving for the magnitude of the tension in the cable yields
mg
T = m(ay + g) = (79 kg)(0.70 m/s2 + 9.80 m/s2) = 8.3 × 102 N
b. The work done by the tension in the cable is
WT = ( T cos θ ) s = (8.3 × 102 N) (cos 0°) (11 m) = 9.1 × 103 J
(6.1)
c. The work done by the person’s weight is
WW = ( mg cos θ ) s = (79 kg) ( 9.8 m/s 2 ) (cos 180°) (11 m) = − 8.5 × 103 J
(6.1)
d. The work-energy theorem relates the work done by the two forces to the
change in the kinetic energy of the person. The work done by the two forces is W = WT +
WW:
WT + WW =
14
243
1
2
mvf2 −
1
2
mv02
W
Solving this equation for the final speed of the person gives
(6.3)
vf =
=
v02 +
2
( WT + WW )
m
( 0 m/s )
2
+
2
( 9.1 × 103 J − 8.5 × 103 J ) =
79 kg
4 m/s
WebAssign Problem 4: A 0.60-kg basketball is dropped out of a window that is 6.1 m
above the ground. The ball is caught by a person whose hands are 1.5 m above the
ground. (a) How much work is done on the ball by its weight? What is the gravitational
potential energy of the basketball, relative to the ground, when it is (b) released and (c)
caught? (d) How is the change (PEf – PE0) in the ball’s gravitational potential energy
related to the work done by its weight?
REASONING The work done by the weight of the basketball is given by Equation 6.1
as W = ( F cos θ ) s , where F = mg is the magnitude of the weight, θ 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 θ = 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.
SOLUTION
a.
The work done by the weight of the basketball is
W = ( F cos θ ) 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
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.
(6.5)
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).
(6.5)
WebAssign Problem 5: A person starts from rest at the top of a large frictionless
spherical surface, and slides into the water below (see the drawing). At what angle θ
does the person leave the surface? (Hint: When the person leaves the surface, the normal
force is zero.)
REASONING AND SOLUTION If air resistance is ignored, the only nonconservative
force that acts on the person is the normal force exerted on the person by the surface.
Since this force is always perpendicular to the direction of the displacement, the work
done by the normal force is zero. We can conclude, therefore, that mechanical energy is
conserved.
1
mv20 +
2
mgh 0 =
1
2
mv 2f + mgh f
(1)
where the final state pertains to the position where
the person leaves the surface. Since the person starts from
rest v0 = 0 m/s. Since the radius of the surface is r,
r
θf
h0 = r, and hf = r cos θf where θf is the angle at which the
person leaves the surface. Equation (1) becomes
mgr =
1
mv 2f
2
+ mg(r cos θ f )
r cos θ f
r
(2)
F
N
In general, as the person slides down the surface, the two
forces that act on him are the normal force FN and the weight
mg. The centripetal force required to keep the person moving
in the circular path is the resultant of FN and the radial
component of the weight, mg cos θ.
mg cos θ
When the person leaves the surface, the normal force is
zero, and the radial component of the weight provides the
centripetal force.
mv 2f
mg cos θ f =
r
⇒
v2f = gr cos θ f
(3)
θ
θ
θ
mg
Substituting this expression for v 2f into Equation (2) gives
1 mg ( r
2
mgr =
Solving for θf gives
cos θ f ) + mg (r cos θ f )
 2
θ f = cos − 1   = 48°
 3
WebAssign Problem 6: Bicyclists in the Tour de France do enormous amounts of work
during a race. For example, the average power per kilogram generated by Lance
Armstrong (m = 75.0) is 6.50 W per kilogram of his body mass. (a) How much work does
he do during a 135-km race in which his average speed is 12.0 m/s? (b) Often, the work
done is expressed in nutritional Calories rather than in joules. Express the work done in
part (a) in terms of nutritional Calories, noting that
nutritional
Calories.
REASONING The work W done is equal to the average power P multiplied by the time
t, or
W = Pt
(6.10a)
The average power is the average power generated per kilogram of body mass
multiplied by Armstrong’s mass. The time of the race is the distance s traveled
divided by the average speed v , or t = s / v (see Equation 2.1).
SOLUTION
a. Substituting t = s / v into Equation 6.10a gives
 135 × 103 m 
W 
 s

6
W = Pt = P   =  6.50
75.0
kg

 = 5.48 × 10 J

v
 12.0 m/s 
kg 

1444
424444
3
P
(
−4
b. Since 1 joule = 2.389 × 10
)
nutritional calories, the work done is
 2.389 × 10− 4 nutritional calories 
W = 5.48 × 106 joules = 5.48 × 106 joules 


1 joule


(
= 1.31× 103 nutritional calories
)
WebAssign Problem 7: Suppose in Figure 6.2 that
of work are done by
the force (
) in moving the suitcase a distance of 50.0 m. At what
angle θ is the force oriented with respect to the ground?
REASONING AND SOLUTION Solving Equation 6.1 for the angle θ , we obtain
 W 
1.10 × 103 J 
−1 
θ = cos − 1 
 = cos 
 = 42.8°
 Fs
 (30.0 N)(50.0 m) 
WebAssign Problem 8: The drawing shows a version of the loop-the-loop trick for a
small car. If the car is given an initial speed of 4.0 m/s, what is the largest value that the
radius r can have if the car is to remain in contact with the circular track at all times?
REASONING AND SOLUTION When the car is at the top of the track the centripetal
force consists of the full weight of the car.
2
mv /r = mg
Applying the conservation of energy between the bottom and the top of the track
gives
2
2
(1/2)mv + mg(2r) = (1/2)mv0
Using both of the above equations
2
v0 = 5gr
so
2
2
2
r = v0 /(5g) = (4.00 m/s) /(49.0 m/s ) = 0.327 m
Practice conceptual problems:
5. A ball has a speed of 15 m/s. Only one external force acts on the ball. After this force
acts, the speed of the ball is 7 m/s. Has the force done positive or negative work? Explain.
REASONING AND SOLUTION The speed of the ball decreases; therefore, the
ball is subjected to an external resistive force. A resistive force always points
opposite to the direction of the displacement of the ball. Therefore, the external force
does negative work.
Using the work-energy theorem, we see that the change in the kinetic energy of the
ball is negative; therefore, the total work done on the ball is negative. We can
conclude, therefore, that the net force did negative work on the ball.
8. The speed of a particle doubles and then doubles again because a net external force
acts on it. Does the net force do more work during the first or the second doubling?
Justify your answer.
REASONING AND SOLUTION The speed of a particle doubles and then doubles
again, because a net external force acts on it. Let m and v0 represent the mass and
initial speed of the particle, respectively. During the first doubling, the change in the
kinetic energy of the particle is
KE f − KE 0 =
1
2
m ( 2v0 ) −
2
1
2
2
mv0 =
3
2
2
mv0
During the second doubling, the change in the kinetic energy of the particle is
KE f − KE 0 =
1
2
m ( 4v0 ) −
2
1
2
m ( 2v0 ) = 6mv02
2
From the work-energy theorem, we know that a change in kinetic energy is equal to
the work done by the net external force. Therefore, more work is done by the net
force during the second doubling.
14. A person is riding on a Ferris wheel. When the wheel makes one complete turn, is
the net work done by the gravitational force positive, negative, or zero? Justify your
answer.
REASONING AND SOLUTION As the person moves downward from the top of
the Ferris wheel, his displacement points downward. Since the person’s weight also
points downward, the work done by gravity is positive. As the person moves upward
from the bottom, his displacement points upward. Since the weight still points
downward, the work done by gravity is negative. The magnitude of the work done in
each half cycle is the same; therefore, the net work done in one revolution is zero.
17. The drawing shows an empty fuel tank about to be released by three different jet
planes. At the moment of release, each plane has the same speed and each tank is at
the same height above the ground. However, the directions of travel are different. In
the absence of air resistance, do the tanks have different speeds when they hit the
ground? If so, which tank has the largest speed and which has the smallest speed?
Explain.
REASONING AND SOLUTION Since each plane has the same speed, the kinetic
energy of each fuel tank will be the same at the instant of release. Since each plane is
at the same height above the ground, each fuel tank must fall through the same
vertical displacement. Therefore, the work done by gravity on each fuel tank is the
same. From the work-energy theorem, each fuel tank will gain the same amount of
kinetic energy during the fall. Therefore, each fuel tank will hit the ground with the
same speed.