217
segments, the velocity graph must consist of line
segments, and the position graph must consist of
parabolas.
x
t
v
t
a
t
bcdef
a
22. We have v
f
2
=2a
.
x, so the distance is proportional
to the square of the velocity. To get up to half the
speed, the ball needs 1/4 the distance, i.e. L/4.
Chapter 4
7. a=
.
v
.
t
, and also a=
F
m
, so
.
t=
.
v
a
=
m
.
v
F
=
(1000kg)(50m/s–20m/s)
3000N
= 10 s
Chapter 5
14. (a)
top spring’s rightward force on connector
...connector’s leftward force on top spring
bottom spring’s rightward force on connector
...connector’s leftward force on bottom spring
hand’s leftward force on connector
Solutions to Selected Problems
...connector’s rightward force on hand
Looking at the three forces on the connector, we see
that the hand’s force must be double the force of
either spring. The value of x-x
o
is the same for both
springs and for the arrangement as a whole, so the
spring constant must be 2k. This corresponds to a
stiffer spring (more force to produce the same
extension).
(b) Forces in which the left spring participates:
hand’s leftward force on left spring
...left spring’s rightward force on hand
right spring’s rightward force on left spring
...left spring’s leftward force on right spring
Forces in which the right spring participates:
left spring’s leftward force on right spring
...right spring’s rightward force on left spring
wall’s rightward force on right spring
...right spring’s leftward force on wall
Since the left spring isn’t accelerating, the total force
on it must be zero, so the two forces acting on it must
be equal in magnitude. The same applies to the two
forces acting on the right spring. The forces between
the two springs are connected by Newton’s third law,
so all eight of these forces must be equal in magni-
tude. Since the value of x-x
o
for the whole setup is
double what it is for either spring individually, the
spring constant of the whole setup must be k/2, which
corresponds to a less stiff spring.
16. (a) Spring constants in parallel add, so the spring
constant has to be proportional to the cross-sectional
area. Two springs in series give half the spring
constant, three springs in series give 1/3, and so on,
so the spring constant has to be inversely propor-
tional to the length. Summarizing, we have k
.
A/L.
(b) With the Young’s modulus, we have k=(A/L)E.The
spring constant has units of N/m, so the units of E
would have to be N/m
2
.
18. (a) The swimmer’s acceleration is caused by the
water’s force on the swimmer, and the swimmer
makes a backward force on the water, which acceler-
ates the water backward. (b) The club’s normal force
on the ball accelerates the ball, and the ball makes a
backward normal force on the club, which deceler-
ates the club. (c) The bowstring’s normal force
accelerates the arrow, and the arrow also makes a
backward normal force on the string. This force on
the string causes the string to accelerate less rapidly
than it would if the bow’s force was the only one
acting on it. (d) The tracks’ backward frictional force
slows the locomotive down. The locomotive’s forward
frictional force causes the whole planet earth to
accelerate by a tiny amount, which is too small to