Chapter 5 solutions
4.
Picture the Problem: You exert a horizontal force that accelerates both your little sister and the sled.
Strategy: Apply Newton’s Second Law to the sled + sister combination, and solve for the mass of your sister.
10.
Solution: 1. Use Newton’s Second Law to find the
total mass of the sled + sister combination:
mtotal 
2. Find the mass of your sister by subtracting:
msister  mtotal – msled  48 kg – 7.4 kg  41 kg
F
120 N

 48 kg
a 2.5 m/s 2
Picture the Problem: The ball is accelerated horizontally in the direction opposite its motion in order to bring it to rest from 92
mi/h over a distance of 0.15 m.
Strategy: Use equation 2-12 to determine the acceleration of the ball, and then use Newton’s Second Law
(equation 5-1) to find the mass from the net force and the acceleration.
Solution: 1. Use equation 2-12 to
find the acceleration of the ball:
vx2  v02x  2ax
v2  v2
02  v02x
v2
a  x 0 x xˆ 
xˆ   0 x xˆ
2 x
2 x
2 x
2. Use equation 5-1 to find the mass:
m
 F  F   2 x    803 N xˆ  
a
2  0.15 m 


  92 mi/h  0.447 m/s mi/h  xˆ 
 v xˆ 
2
0x
2
m  0.14 kg
18.
Picture the Problem: The free body diagrams for the car and the trailer is
shown at right. The diagram assumes there is no friction.
Strategy: In order to determine the forces acting on an object, you must consider
only the forces acting on that object and the motion of that object alone. For the
trailer there is only one force F1 exerted on it by the car, and it has the same
acceleration (1.85 m/s2) as the car. For the car there are two forces acting on it,
the engine F2 and the trailer F1 . Apply Newton’s Second and Third Laws as
appropriate to find the requested forces.
 m a   560 kg  1.85 m/s 2  xˆ  1.0 kN  xˆ
Solution: 1. (a) Write Newton’s Second Law for the trailer:
F  F
2. (b) Newton’s Third Law states that the force the trailer
exerts on the car is equal and opposite to the force the car
exerts on the trailer:
F1   1.0 kN  xˆ
3. (c) Write Newton’s Second Law for the car:
 F  M a  1400 kg  1.85 m/s  xˆ   2.6 kN  xˆ
1
2
20.
Picture the Problem: The force pushes on box 1 in the manner indicated by
the figure at right.
Strategy: The boxes must each have the same acceleration, but because they
have different masses the net force on each must be different. These
observations allow you to use Newton’s Second Law for each individual box
to determine the magnitudes of the contact forces. First find the acceleration
of all the boxes and then apply equation 5-1 to find the contact forces.
Solution: 1. (a) The 7.50 N force
accelerates all the boxes together:
F  (m1  m2  m3 )a  a 
a
2. Write Newton’s Second Law for the first box:
F
m1  m2  m3
7.50 N
 0.798 m/s 2
1.30 kg  3.20 kg  4.90 kg
F  F  F
c12
 m1 a
Fc12  F  m1 a  7.50 N  1.30 kg   0.798 m/s 2   6.46 N
3. (b) Write Newton’s Second Law for the third box:
27.
F  F
c23
 m3 a   4.90 kg   0.798 m/s 2   3.91 N
Picture the Problem: The cart is pushed partly into the incline and partly up the
incline by the pushing force F , as shown in the figure at right.
Strategy: Write Newton’s Second Law for the x direction, where x̂ points up
the incline and parallel to it. Solve the resulting equation for the magnitude of
F.
Solution: The component of the force pushing up the incline
is F cos  and the component of the weight pushing down
the incline is mg sin  :
F
x
 F cos   mg sin   ma
2
ma  mg sin   7.5 kg  1.41   9.81 m/s  sin13
F

cos 
cos13
F  28 N
30.
Picture the Problem: The two teenagers pull on the sled in the directions indicated by the figure at right.
Strategy: Write Newton’s Second Law in the x direction (parallel to a ) in order to find the acceleration of the sled.
Solution: Write Newton’s Second Law
in the x direction:
41.
F
 2 F cos 35  57 N   msled  mchild  ax
2 F cos 35  57 N
ax 
msled  mchild
2  55 N  cos 35  57 N
ax 
 1.5 m/s 2
19  3.7 kg
x
Picture the Problem: The elevator accelerates up and down, changing your apparent
weight Wa. A free body diagram of the situation is depicted at right.
Strategy: There are two forces acting on you: the applied force F  Wa of the scale acting
upward and the force of gravity W acting downward. The force Wa represents your
apparent weight because it is both the force the scale exerts on you and the force you exert
on the scale. Use Newton’s Second Law together with the known force Wa acceleration to
determine the acceleration a.
Solution: 1. The direction of acceleration is downward. A downward acceleration results
in an apparent weight less than the actual weight.
F
2. Use Newton’s Second Law together
with the known forces to determine the
magnitude of the acceleration a.
50.
y
 Wa  W  ma
a 
Wa  W
W W
121  142 lb
 a

 9.81 m/s2   1.5 m/s2
m
W g
142 lb
Picture the Problem: The free body diagram of the lawn mower is shown at right.
Strategy: Write Newton’s Second Law in the vertical direction to determine the
normal force.
Solution: 1. (a) Use
Newton’s Second
Law to find N:
F
 N  F sin   mg  ma y  0
N  F sin   mg
  219 N  sin 35  19 kg   9.81 m/s 2 
y
θ
N  310 N  0.31 kN
2. (b) If the angle between the handle and the horizontal is increased, the normal force exerted by the lawn will increase because
it must still balance the weight plus a larger downward force than before.