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CHAPTER 5
PROBLEM SOLUTIONS
Problem
4. A car leaves the road traveling at 110 km/h and hits a tree, coming to a complete stop in 0.14 s. What average force does
a seatbelt exert on a 60-kg passenger during this collision?
Solution
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
a
av
=
(0
!
v
0
)
=
t
, and the average net force on the
passenger, while coming to rest, is
F
av
=
ma
av
=
!
m
v
0
=
t
=
!
(60 kg)(110
=
3.6)(m/s)
=
(0.14 s)
=
!
13.1 kN
, 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.)
Problem
7. Object A accelerates at 8.1 m/s
2
when a 3.3-N force is applied. Object B accelerates at 2.7 m/s
2
when the same force is
applied. (a) How do the masses of the two objects compare? (b) If A and B were stuck together and accelerated by the
3.3-N force, what would be the acceleration of the composite object?
Solution
In this idealized one-dimensional situation, the applied force of
F
=
3.3 N
is the only force acting. (a) When applied to
either object, Newton’s second law gives
F
=
m
A
a
A
and
F
=
m
B
a
B
, so
m
B
=
m
A
=
a
A
=
a
B
=
(8.1 m/s
2
)
=
(2.7 m/s
2
)
=
3.
(For constant net force, mass is inversely proportional to acceleration.) (b) When
F
is applied to the combined object,
F
=
(
m
A
+
m
B
)
a
.
Since
F
=
m
A
a
A
, and
m
B
=
3
m
A
,
one finds
a
=
F
=
(
m
A
+
m
B
)
=
m
A
a
A
=
4
m
A
=
1
4
(8.1 m/s
2
)
=
2.03 m/s
2
. (Note: It was not necessary to calculate the masses, which are
m
A
=
(3.3 N)
=
2
)
=
0.407 kg
, and
m
B
=
(3.3 N)
=
(2.7 m/s
2
)
=
1.22 kg.)
Problem
9. By how much does the force required to stop a car increase if the initial speed is doubled and the stopping distance
remains the same?
Solution
The average net force on a car of given mass is proportional to the average acceleration,
F
av
»
a
av
.
To stop a car in a given
distance,
(
x
!
x
0
),
a
av
=
(0
!
v
0
2
)
=
2(
x
!
x
0
),
so
F
av
»
v
0
2
.
Doubling
v
0
quadruples the magnitude of
F
av
, a fact that is
important to remember when driving at high speeds.
Problem
11. The maximum braking force of a 1400-kg car is about 8.0 kN. Estimate the stopping distance when the car is traveling
(a) 40 km/h; (b) 60 km/h; (c) 80 km/h; (d) 55 mi/h.
Solution
The maximum braking acceleration is
a
=
F
=
m
=
!
8
.
0
"
10
3
N
=
1400 kg
=
!
5
.
71 m
/
s
2
.
(We expressed the braking force as
negative because it is opposite to the direction of motion.) (a) On a straight horizontal road, a car traveling at a velocity of
v
0
x
=
40 km/h
can stop in a minimum distance found from Equation 2-11 and the maximum deceleration just calculated:
x
!
x
0
=
!
v
0
x
2
=
2
a
=
!
(40 m
=
3.6 s)
2
=
2(
!
5.71 m/s
2
)
=
10.8 m.
For the other initial velocities, the stopping distance is
(b) 24.3 m, (c) 43.2 m, and (d) 52.9 m.

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