Physics 107 HOMEWORK ASSIGNMENT #4
Cutnell & Johnson, 7th edition
Chapter 4: Problems 15, 44, 58, 106, 108
*15 A duck has a mass of 2.5 kg. As the duck paddles, a force of 0.10 N acts on it in a direction
due east. In addition, the current of the water exerts a force of 0.20 N in a direction of 52° south
of east. When these forces begin to act, the velocity of the duck is 0.11 m/s in a direction due
east. Find the magnitude and direction (relative to due east) of the displacement that the duck
undergoes in 3.0 s while the forces are acting.
*44 Refer to Multiple-Concept Example 10 for help in solving problems like this one. An ice
skater is gliding horizontally across the ice with an initial velocity of +6.3 m/s. The coefficient of
kinetic friction between the ice and the skate blades is 0.081, and air resistance is negligible.
How much time elapses before her velocity is reduced to +2.8 m/s ?
*58 The person in the drawing is standing on crutches. Assume that the force
exerted on each crutch by the ground is directed along the crutch, as the force
vectors in the drawing indicate. If the coefficient of static friction between
a crutch and the ground is 0.90, determine the largest angle
the crutch can have just before it begins to slip on the floor.
that
*106 Jupiter is the largest planet in our solar system, having a mass
and radius that are, respectively, 318 and 11.2 times that of earth. Suppose
that an object falls from rest near the surface of each planet and that the acceleration due to
gravity remains constant during the fall. Each object falls the same distance before striking the
ground. Determine the ratio of the time of fall on Jupiter to that on earth.
**108 As part a of the drawing shows, two blocks are connected by a rope that passes over a set
of pulleys. One block has a weight of 412 N, and the other has a weight of 908 N. The rope and
the pulleys are massless and there is no friction. (a) What is the acceleration of the lighter block?
(b) Suppose that the heavier block is removed, and a downward force of 908 N is provided by
someone pulling on the rope, as part b of the drawing shows. Find the acceleration of the
remaining block. (c) Explain why the answers in (a) and (b) are different.
15. REASONING
Equations 3.5a
 x  v0 xt  12 axt 2 
and 3.5b
 y  v0 yt  12 ayt 2 
give the
displacements of an object under the influence of constant accelerations ax and ay. We can
add these displacements as vectors to find the magnitude and direction of the resultant
displacement. To use Equations 3.5a and 3.5b, however, we must have values for ax and ay.
We can obtain these values from Newton’s second law, provided that we combine the given
forces to calculate the x and y components of the net force acting on the duck, and it is here
that our solution begins.
SOLUTION Let the directions due east and due north, respectively, be the +x and +y
directions. Then, the components of the net force are
Fx  0.10 N   0.20 N  cos 52  0.2231 N
Fy  –  0.20 N  sin 52  –0.1576 N
According to Newton’s second law, the components of the acceleration are
ax 
ay 
Fx
m
Fy
m

0.2231 N
 0.08924 m/s 2
2.5 kg

–0.1576 N
 –0.06304 m/s 2
2.5 kg
From Equations 3.5a and 3.5b, we now obtain the displacements in the x and y directions:


x  v0 xt  12 axt 2   0.11 m/s  3.0 s   12 0.08924 m/s 2  3.0 s   0.7316 m

2

y  v0 y t  12 a y t 2   0 m/s  3.0 s   12 –0.06304 m/s 2  3.0 s   –0.2837 m
2
The magnitude of the resultant displacement is
r  x2  y 2 
 0.7316 m 2   –0.2837 m 2
 0.78 m
The direction of the resultant displacement is
 0.2837 m 
  21 south of east
 0.7316 m 
  tan –1 
44. REASONING Let us assume that the skater is moving horizontally along the +x axis. The
time t it takes for the skater to reduce her velocity to vx = +2.8 m/s from v0x = +6.3 m/s can
be obtained from one of the equations of kinematics:
vx  v0 x  ax t
(3.3a)
The initial and final velocities are known, but the acceleration is not. We can obtain the
acceleration from Newton’s second law  Fx  max , Equation 4.2a  in the following
manner. The kinetic frictional force is the only horizontal force that acts on the skater, and,
since it is a resistive force, it acts opposite to the direction of the motion. Thus, the net force
in the x direction is Fx   f k , where fk is the magnitude of the kinetic frictional force.
Therefore, the acceleration of the skater is ax  Fx m   f k / m .
The magnitude of the frictional force is f k  k FN (Equation 4.8), where k is the
coefficient of kinetic friction between the ice and the skate blades and FN is the magnitude
of the normal force. There are two vertical forces acting on the skater: the upward-acting
normal force FN and the downward pull of gravity (her weight) mg. Since the skater has no
vertical acceleration, Newton's second law in the vertical direction gives (taking upward as
the positive direction) Fy  FN  mg  0 . Therefore, the magnitude of the normal force is
FN  mg and the magnitude of the acceleration is
ax 
 f k  k FN  k m g


  k g
m
m
m
SOLUTION
Solving the equation vx  v0 x  ax t for the time and substituting the expression above for
the acceleration yields
t
vx  v0 x
ax

vx  v0 x
 k g

2.8 m/s  6.3 m/s
 4.4 s
  0.081  9.80 m/s 2 
58. REASONING The diagram at the right shows the force F that the
ground exerts on the end of a crutch. This force, as mentioned in the
F
statement of the problem, acts along the crutch and, therefore, makes
an angle  with respect to the vertical. The horizontal and vertical
components of this force are also shown. The horizontal component,
F sin , is the static frictional force that prevents the crutch from F sin 
slipping on the floor, so fs  F sin  . The largest value that the static
frictional force can have before the crutch begins to slip is then given
F cos 

by fsMAX  F sin  MAX . We also know from Section 4.9 (see Equation 4.7) that the
maximum static frictional force is related to the magnitude FN of the normal force by
fsMAX  s FN , where s is the coefficient of static friction. These two relations will allow
us to find  MAX .
SOLUTION The magnitude of the maximum static frictional force is given by
fsMAX  s FN . But, as mentioned in the Reasoning section, fsMAX is also the horizontal
component of the force F, so
fsMAX  F sin  MAX . The vertical component of F,
F cos MAX is the magnitude FN of the normal force that the ground exerts on the
crutch. Thus, we have
fsMAX  s
FN
F cos MAX
F sin  MAX
The force F can be algebraically eliminated from this equation, leaving
sin  MAX
 s
cos  MAX
or
tan MAX  s
The maximum angle that a crutch can have is
 MAX  tan 1  s   tan 1  0.90   42
106. REASONING AND SOLUTION
acceleration, it follows that
Since both motions are characterized by constant
yJ
yE

1 a t2
2 J J
1 a t2
2 E E
where the subscripts designate those quantities that pertain to Jupiter and Earth. Since both
objects fall the same distance, the above ratio is equal to unity. Solving for the ratio of the
times yields
tJ
tE

aE
GM E / RE2 RJ


aJ
GM J / RJ2 RE
ME
1
 11.2 
 0.628
MJ
318
108. REASONING AND SOLUTION
a. The rope exerts a tension, T, acting upward on each block. Applying Newton's second
law to the lighter block (block 1) gives
T – m1g = m1a
Similarly, for the heavier block (block 2)
T – m2g = – m2a
Subtracting the second equation from the first and rearranging yields
 m – m1 
a 2
g  3.68 m/s 2
 m  m 
1
 2
b. The tension in the rope is now 908 N since the tension is the reaction to the applied force
exerted by the hand. Newton's second law applied to the block is
T – m1g = m1a
Solving for a gives
a
 908 N  – 9.80 m/s 2  11.8 m/s 2
T
–g
m1
42.0 kg
c. In the first case, the inertia of BOTH blocks affects the acceleration whereas, in the
second case, only the lighter block's inertia remains.