Physics 20 Homework Set 4
Chapter 4 Problems 6,10,11,13,14,24,27,29,32,39
v 2  v02  320 m s   0

2  x 
2  0.82 m 
2
4.6
The acceleration of the bullet is given by a 
  320 m s 2 
Then, F  ma  5.0  103 kg 
  3.1  102 N
2
0.82
m
 
 

4.10

(a) Choose the positive y-axis in the forward direction. We
resolve the forces into their components as
Force
400 N
450 N
Resultant
x-component
200 N
–78.1 N
y-component
346 N
443 N
Fx  122 N
Fy  790 N
The magnitude and direction of the resultant force is
FR 
 Fx    Fy 
2
2
 F
 799 N ,   tan 1  x
 Fy


  81.2 to right of y-axis.

Thus, FR  799 N at 81.2 to the right of the forward direction
(b) The acceleration is in the same direction as FR and has magnitude
a
FR
799 N

 0.266 m s2
m 3 000 kg
4.11
Starting with v0 y  0 and falling 30 m to the ground, the velocity of the ball just
before it hits is


v1   v02 y  2a y y   0  2 9.80 m s 2  30 m   24 m s
On the rebound, the ball has v y  0 after a displacement y  20 m . Its velocity
as it left the ground must have been


v2   v 2y  2a y y   0  2 9.80 m s 2  20 m   20 m s
Thus, the average acceleration of the ball during the 2.0-ms contact with the
ground was
aav 
v2  v1 20 m s   24 m s 

 2.2  104 m s2
3
t
2.0  10 s
The resultant force acting on the ball during this time interval must have been


F  ma   0.50 kg  2.2  104 m s2  1.1  104 N
or
4.13
F  1.1  104 N upward
From Fx  0 , T1 cos30.0  T2 cos60.0  0
or
T2  1.73 T1
(1)
Then Fy  0 becomes
T1 sin 30.0  1.73 T1  sin 60.0  150 N  0
which gives T1  75.0 N in the right side cable
Finally, Equation (1) above gives T2  130 N in the left side cable
4.14
If the hip exerts no force on the leg, the system must be
in equilibrium with the three forces shown in the freebody diagram.
Thus Fx  0 becomes
w2 cos  110 N  cos40
(1)
From Fy  0 , we find
w2 sin   220 N  110 N  sin 40
(2)
Dividing Equation (2) by Equation (1) yields
 220 N  110 N  sin 40 
  61
110
N
cos
40





  tan 1 
Then, from either Equation (1) or (2), w2  1.7  102 N
4.24
First consider the block moving along the
horizontal. The only force in the direction
of movement is T. Thus,
Fx  ma x  T   5.00 kg  a
(1)
Next consider the block which moves
vertically. The forces on it are the tension
T and its weight, 98.0 N.
Fy  ma y  98.0 N  T  10.0 kg  a
(2)
Note that both blocks must have the same magnitude of acceleration. Equations
(1) and (2) can be solved simultaneously to give.
a  6.53 m s2 , and T  32.7 N
4.27
Choose the +x direction to be horizontal and forward with the +y vertical and
upward. The common acceleration of the car and trailer then has components of
a x  2.15 m s2 and a y  0 .
(a) The net force on the car is horizontal and given by
Fx car  F  T  mcar ax  1000 kg   2.15 m s2  
2.15 103 N forward
(b) The net force on the trailer is also horizontal and given by
 Fx trailer  T  mtrailer ax   300 kg   2.15

m s2  645 N forward
(c) Consider the free-body diagrams of the car and trailer. The only horizontal
force acting on the trailer is T  645 N forward , and this is exerted on the
trailer by the car. Newton’s third law then states that the force the trailer
exerts on the car is 645 N toward the rear
(d) The road exerts two forces on the car. These are F and nc shown in the freebody diagram of the car.
From part (a),
Also,  Fy 
car
F  T  2.15 103 N   2.80 103 N
 nc  wc  mcar a y  0 , so nc  wc  mcar g  9.80 103 N
The resultant force exerted on the car by the road is then
Rcar  F 2  nc2 
 2.80 10 N   9.80 10 N 
3
2
3
2
 1.02  104 N
n 
at   tan 1  c   tan 1  3.51  74.1 above the horizontal and forward.
F
Newton’s third law then states that the resultant force exerted on the road
by the car is
1.02  104 N at 74.1 below the horizontal and rearward
4.29
When the block is on the verge of moving, the static friction force has a
magnitude f s   f s max  s n .
Since equilibrium still exists and the applied force is 75 N, we have
Fx  75 N  f s  0 or  f s max  75 N
In this case, the normal force is just the weight of the crate, or n  mg . Thus, the
coefficient of static friction is
s 
 f s max
n

 f s max
mg

75 N
 0.38
 20 kg   9.80 m s2 
After motion exists, the friction force is that of kinetic friction, f k  k n
Since the crate moves with constant velocity when the applied force is 60 N, we
find that Fx  60 N  f k  0 or f k  60 N . Therefore, the coefficient of kinetic
friction is
k 
4.32
(a)
ax 
fk
f
60 N
 k 
 0.31
n mg  20 kg   9.80 m s2 
vx  v0 x 6.00 m s  12.0 m s

  1.20 m s2
t
5.00 s
(b) From Newton’s second law, Fx   f k  ma x , or f k  ma x .
The normal force exerted on the puck by the ice is n  mg , so the coefficient
of friction is
k 
(c)



2
f k m 1.20 m s

 0.122
n
m 9.80 m s2

 v  v   6.00 m s  12.0 m s 
x   vx av t   x 0 x  t  
 5.00 s   45.0 m
2
 2  

4.39
The acceleration of the system is found from
1
1
2
y  v0 y t  at 2 , or 1.00 m  0  a 1.20 s 
2
2
which gives a  1.39 m s2
Using the free body diagram of m2 , the second law gives
5.00 kg   9.80
or


m s2  T  5.00 kg  1.39 m s2

T  42.1 N
Then applying the second law to the horizontal motion of m1


42.1 N  f  10.0 kg  1.39 m s2 , or f  28.2 N
Since n  m1 g  98.0 N , we have k 
f 28.2 N

 0.287
n 98.0 N