Chapter 8 – Student Solutions Manual
5. The potential energy stored by the spring is given by
Uk
x
=
1
2
2
, where
k
is the spring
constant and
x
is the displacement of the end of the spring from its position when the
spring is in equilibrium. Thus
k
U
x
==
=
×
2
225
0 075
89 10
2
2
3
J
m
Nm
b
g
b
g
.
..
9. We neglect any work done by friction. We work with SI units, so the speed is
converted:
v
= 130(1000/3600) = 36.1 m/s.
(a) We use Eq. 8-17:
K
f
+
U
f
=
K
i
+
U
i
with
U
i
= 0,
U
f
=
mgh
and
K
f
= 0. Since
Km
v
i
=
1
2
2
, where
v
is the initial speed of the truck, we obtain
22
2
2
1
(36.1m/s)
66.5 m
2
2
2(9.8 m/s )
v
mv
mgh
h
g
=⇒
=
=
=
.
If
L
is the length of the ramp, then
L
sin 15° = 66.5 m so that
L
= 66.5/sin 15° = 257 m.
Therefore, the ramp must be about 2.6
×
10
2
m long if friction is negligible.
(b) The answers do not depend on the mass of the truck. They remain the same if the
mass is reduced.
(c) If the speed is decreased,
h
and
L
both decrease (note that
h
is proportional to the
square of the speed and that
L
is proportional to
h
).
11. (a) If
K
i
is the kinetic energy of the flake at the edge of the bowl,
K
f
is its kinetic
energy at the bottom,
U
i
is the gravitational potential energy of the flake-Earth system
with the flake at the top, and
U
f
is the gravitational potential energy with it at the bottom,
then
K
f
+
U
f
=
K
i
+
U
i
.
Taking the potential energy to be zero at the bottom of the bowl, then the potential energy
at the top is
U
i
=
mgr
where
r
= 0.220 m is the radius of the bowl and
m
is the mass of the
flake.
K
i
= 0 since the flake starts from rest. Since the problem asks for the speed at the
bottom, we write
1
2
2
mv
for
K
f
. Energy conservation leads to
WF
d
m
g
h
m
g
L
gg
=⋅
=
=
−
G
G
(c
o
s
)
1
θ
.
The speed is
.
0
8
m
vg
r
/
s
.

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*Sign up*(b) Since the expression for speed does not contain the mass of the flake, the speed would
be the same, 2.08 m/s, regardless of the mass of the flake.
(c) The final kinetic energy is given by
K
f
=
K
i
+
U
i
–
U
f
. Since
K
i
is greater than before,
K
f
is greater. This means the final speed of the flake is greater.
31. We refer to its starting point as
A
, the point where it first comes into contact with the
spring as
B
, and the point where the spring is compressed |
x
| = 0.055 m as
C
. Point
C
is
our reference point for computing gravitational potential energy. Elastic potential energy
(of the spring) is zero when the spring is relaxed. Information given in the second
sentence allows us to compute the spring constant. From Hooke's law, we find
k
F
x
==
=
×
270 N
0.02 m
1.35 10 N m
4
.
(a) The distance between points
A
and
B
is
G
F
g
and we note that the total sliding distance
A
+
x
is related to the initial height
h
of the block (measured relative to
C
) by
h
x
A
+
=
sin
θ
where the incline angle
is 30°. Mechanical energy conservation leads to
KU
mgh
kx
AACC
+
=
+
+=
+
0
1
2
0
2
which yields
h
kx
mg
×
=
2
2
2
2
212
98
1.35 10 N m 0.055 m
kg
m s
0.174 m
4
ch
bg
b
g
.

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