4.1
FORCE AND MOTION
EXERCISES
Section 4.2 Newton’s First and Second Laws
13.
INTERPRET
We interpret this as a problem involving the application of Newton’s second law. The object under
consideration is the train and the physical quantity of interest is the net force acting on it.
DEVELOP
The net force can be found by using Equation 4.3,
r
Fm
a
net
=
.
EVALUATE
Using Equation 4.3, the magnitude of the force acting on the train is found to be
a
net
kg
m s
N
==
×
=
×
(.
)
/ )
.
15 10
25
375 10
62 6
ASSESS
The result is reasonable, since by definition, one newton is the force required to accelerate a 1-kg mass at
the rate of 1 m/s
2
.
14.
Ignoring the probable presence of other forces, we can apply Equation 4.3 to find
(a)
aF
m
==×
/(
.
)
/
12 10
5
N
)
.
/ ,
61 10
197
42
×=
kg
m s
and
(b)
a
=×
×
=
)
/
)
.
/ .
146 10
821
56
2
Nk
g
c
m
s
15.
INTERPRET
We interpret this as a problem involving the application of Newton’s second law. The object under
consideration is the airplane and the physical quantity of interest is the plane’s mass.
DEVELOP
We shall assume that the runway is horizontal (so that the vertical force of gravity and the normal
force of the surface cancel) and neglect aerodynamic forces (which are small just after the plane begins to move).
Then the net force equals the engine’s thrust and is parallel to the acceleration. The plane’s mass can be found by
using Equation 4.3,
r
r
a
net
=
.
EVALUATE
Using Equation 4.3, the mass of the plane is found to be
m
F
a
×
net
N
ms
kg
11 10
72
153 10
4
2
3
.
./
.
ASSESS
First, the units are consistent since
11
N
kg m/s
2
=⋅
. The result is reasonable, since by definition, one
newton is the force required to accelerate a 1-kg mass at the rate of 1 m/s
2
.
16.
Assume that the seatbelt holds the passenger firmly to the seat, so that the passenger also stops in 0.14 s without
incurring any secondary impact. Then the passenger’s average acceleration is
av
t
av
=−
()
/
,
0
0
and the average net
force on the passenger, while coming to rest, is
a
m
v
t
av
av
kg
m s
−=
−
0
60
110 3 6
/
(
)(
/ . )( / )/
)
.
014
131
sk
N
,
or about 1.5 tons. (Here, we used the one-dimensional form of Newton’s second law. The
minus sign indicates that the force is opposite to the direction of the initial velocity. It is reasonable to assume that
this is a component of the force exerted by the seatbelt.)
17.
INTERPRET
In this problem we want to find the relationship between the initial speed of the car and the force
required to stop it.
DEVELOP
From Equation 4.3, we see that the net force on a car of given mass is proportional to the acceleration,
Fa
net
;
. We can then relate the three quantities, displacement, velocity, and acceleration, by Equation 2.11,
vv a
x
x
2
0
2
0
2
=+
−
.
EVALUATE
To stop a car in a given distance,
,
xx
−
0
the acceleration is
a
vv
v
−
−
−
−
2
0
2
0
0
2
0
22
() ()
.
Therefore, we see that
Fv
net
;
0
2
. Doubling
v
0
quadruples the magnitude of
F
net
.