Chapter 4 Homework Problem Solutions (#1, 2, 3, 6, 7, 9, 24)
1.
Use Newton’s second law to calculate the force.
 F  ma   60.0 kg  1.25

m s 2  75.0 N
2.
Use Newton’s second law to calculate the mass.
 F 265 N
 F  ma  m  a  2.30 m s2  115 kg
3.
Use Newton’s second law to calculate the tension.
F  F
T
6.


 ma   960 kg  1.20 m s 2  1.15  103 N
Find the average acceleration from Eq. 2-2. The average force on the car is found from Newton’s second law.
v  v0 0  26.4 m s
 0.278 m s 
v  0 v0   95 km h  
 26.4 m s aavg 

 3.30 m s 2

t
8.0 s
 1km h 


Favg  maavg  1100 kg  3.3 m s 2  3.6  103 N
The negative sign indicates the direction of the force, in the opposite direction to the initial velocity.
7. The average force on the pellet is its mass times its average acceleration. The average acceleration is
found from Eq. 2-11c. For the pellet, v0  0 , v  125m s , and x  x0  0.800 m .
aavg
125 m s   0


 9770
2  x  x0 
2  0.800 m 
2
v 2  v02


m s2

Favg  maavg  7.00  103 kg 9770 m s 2  68.4 N
9.
The problem asks for the average force on the glove, which in a direct calculation would require knowledge about
the mass of the glove and the acceleration of the glove. But no information about the glove is given. By
Newton’s 3rd law, the force exerted by the ball on the glove is equal and opposite to the force exerted by the glove
on the ball. So calculate the average force on the ball, and then take the opposite of that result to find the average
force on the glove. The average force on the ball is its mass times its average acceleration. Use Eq. 2-11c to find
the acceleration of the ball, with v  0 , v0  35.0 m s , and x  x0  0.110 m . The initial direction of the ball is
the positive direction.
aavg 
v 2  v02
2  x  x0 

0   35.0 m s 
2  0.110 m 

2
 5568 m s 2

Favg  maavg   0.140 kg  5568 m s 2  7.80  102 N
Thus the average force on the glove was 780 N, in the direction of the initial velocity of the ball.
solution
24.
The net force in each case is found by vector addition with components.
FNet x   F1  10.2 N FNet y   F2  16.0 N
(a)
 10.2    16.0 
FNet 
2
2
 19.0 N
  tan
1
16.0
10.2
o
The actual angle from the x-axis is then 237.5 .
ax = FNetx / m = -10.2 N / 27.0 kg = -.378 m/sec2
ay = FNety / m = -16.0 N / 27.0 kg = -.593 m/sec2
a
FNet
m

19.0 N
27.0 kg
 57.5
F1

o
F2
Fnet
 0.703m s 2 at 237 o
F2
Fnet

(b)
FNet x  F1 cos 30  8.83 N
o
FNet 
a
FNet
FNet y  F2  F1 sin 30  10.9 N
 8.83 N   10.9 N 
2

14.0 N
30o
o
2
 14.0 N
  tan 1
10.9
8.83
 0.520 m s 2 at 51.0o
m
27.0 kg
the individual components of acceleration are
ax = FNetx / m = 8.83 N / 27.0 kg = .327 m/sec2
ay = FNety / m = 10.9 N / 27.0 kg = .404 m/sec2
F2
Fnet

30o
F1
F1
 51.0 o