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Chapter 3
Newton’s Laws
Practice Exercise
Solutions
1. Of the following, the one most closely related to inertia is:
A) weight
B) acceleration
C) mass
D) force
2. If your mass is 75kg, the magnitude of the Earth’s
gravitational pull on you is:
A) 132N
B) 588N
C) 600N
D) 735N
FG  w  mg  (75kg)(9.8 m2 )  735N
s
Magnitude  w   735N  735N
3. The net force required to keep a
10kg object moving at a constant
acceleration of 5 m2 is :
s
A) zero
B) 2N
C) 10N
D) 50N
kg  m
m
Fnet = ma = (10kg)(5 2 )  50 2  50N
s
s
4. If you observe an object which is not moving you can conclude
that no forces are acting on it.
You can only conclude that no NET FORCE is
A) True
acting on it, there may be any number of forces
B) False
acting on the object…we can only conclude that the
sum of the forces is zero.
5. Constant acceleration is to net force as constant velocity is to:
A) weight
B) inertia
C) free-fall
D) work
Motion at a constant acceleration is caused by a net force.
Motion at a constant velocity is caused by inertia.
6. Action and reaction forces never cancel each other because:
A) they are not equal
B) they act on different objects
C) they are not opposite to each other
D) all of these reasons
7. If the net force acting on an object is doubled and the mass of
the object is kept constant, the acceleration of the object will be
multiplied by:
The acceleration of an object is directly
A) 1/2
proportional to the net force acting on it. If the net
B) 2
force is multiplied by some factor and the mass is
C) 1/4
held constant the acceleration will be multiplied by
D) 4
the same factor. Doubling the net force will double
the acceleration.
The acceleration is inversely proportional to the object’s mass.
Multiplying the mass by some factor and keeping the net force
constant will multiply the acceleration by the reciprocal of that
factor. Doubling the mass will multiply the acceleration by 1/2.
8. A 15kg object is initially moving at 100m/s to the right. The net
force required to stop the object in 5s is:
A) 60N
B) -60N
C) 300N
D) -300N
Accelerated Motion
m
0

100
vf  vi
s
d?
a?

a
5s
t
v?
t  5s
 20 m2
s
m
m
=
15kg
v i  100 s
Fnet = ? F = ma = (15kg)(20 m )
vf  0
net
2
s
 300N
9. A 20kg box is being pulled to the right along a level floor by a
force of 90N. If the acceleration of the box is 2.5m/s2, the force of
kinetic friction between the box and the floor must be:
A) -20N
a = 2.5 m2 
B) -30N
s
Kinetic
Fapp  90N
C) -40N
Friction 20kg
D) -50N
f
Fnet
Fnet
If this was the ONLY force
the acceleration would be:
a = 90N  4.5 m2
20kg
s
m
 ma  (20kg)(2.5 2 )  50N
s
 Fapp  f
f  Fnet  Fapp  50N  90N  40N
10. The upward force required to lift an 11kg object at a constant
velocity is:
Fapp
A) 107.8N
B) 215.6N
C) 312.4N
D) zero
11kg
FG
v constant  a = 0  Fnet = 0
Fnet = Fapp + FG
Fapp
m

(11kg)(9.8
= FG  mg
2 )
s
 107.8N
11. A 7.5kg object is acted upon by two horizontal forces: F1 is 35N
to the left, F2 is 80N to the right. The acceleration of the object is:
A) 7.7 m2
s
B)  7.7 m2
s
C) 6.0 m2
s
D) 3.0 m2
s
F1  35N
7.5kg
F2  80N
Fnet  F1  F2  (35N)  (80N
 45N
Fnet
m
45N

6
a  m  7.5kg
s2
12. A person standing at the bottom of a 30m high tower throws a
2kg ball straight up with an initial velocity of 25m/s. Will the ball
reach the top of the tower?
A) Yes
B) No
Free-Fall
d?
m  2kg
v?
Fnet  FG
v i  25 m
s
v f  0@ max .alt.
a  g  9.8 m2
s
t?
2
vf

2
vi
 2ad
v 2f  v 2i
d
2a
2
0  (25 m
)
s
d
2(9.8 m2 )
s
d  31.9m  30m
13. A .2kg helium balloon is acted upon by a buoyant force of 4N.
If released from rest the time required for the balloon to reach an
altitude of 200m would be:
F  4N
A) 6.3s
B) 14.1s
C) 18.5s
D) 22.6s
Accelerated Motion
B
.2kg
FG  w  mg
 (.2kg)(9.8 m2 )
s
 1.96N
d  200m
v?
vi  0
vf  ?
a  ? 10.2 m2
s
t?
m  .2kg
Fnet  ?2.04N
Fnet  FB  FG  4N  (1.96N)  2.04N
Fnet  2.04N  10.2 m
a m
.2kg
s2
0 1 2
d  v i t  2 at
2
1
d  2 at
t
2d 
a
2(200m)  6.3s
10.2 m2
s
14. Ball A is dropped from a cliff 100m high. At the same time Ball
B is thrown horizontally from another cliff 100m high. The paths
followed by the two balls are shown below.
The ball that reaches the ground first will be:
A) Ball A
B) Ball B
C) they will reach the ground at the same time
D) the initial velocity of Ball B must be known