Chapter 8
Problem Solutions
Problem
2. Now take Fig. 8-26 to lie in a vertical plane, and find the work done by the gravitational force as an object moves from
point 1 to point 2 over each of the paths shown.
Solution
Take the origin at point 1 in Fig. 8-26 with the
x
-axis horizontal to the right and the
y
-axis vertical upward. The gravitational
force on an object is constant,
F
g
=
!
mg
ˆ
j
,
while the paths are (a)
d
r
=
ˆ
j
dy
for
x
=
0
and
0
!
y
!
l
,
followed by
d
r
=
î
dx
for
y
=
l
and
0
!
x
!
l
,
and (b)
d
r
=
î
dx
+
ˆ
j
dy
=
(
î
+
ˆ
j
)
dy
,
for
0
!
y
!
l
(since
x
=
y
along this path).
The work done by gravity (Equation 7-11) is
W
g
(
a
)
=
F
g
!
d
r
=
"
(
#
mg
ˆ
j
)
!
ˆ
j
dy
0
$
"
+
(
#
mg
ˆ
j
)
!
î
dx
0
$
"
=
#
mg
dy
+
0
=
#
mg
l
,
0
$
"
and
W
g
(
b
)
=
(
!
mg
ˆ
j
)
"
!
(
î
î
+
ˆ
j
0
l
#
)
dy
=
!
mg
dy
0
l
#
=
!
mg
l
.
Of course, these must be the same because gravity is a conservative force.
Problem
10. A 60-kg hiker ascending 1250-m-high Camel’s Hump mountain in Vermont has potential energy
!
2.4
"
10
5
J;
the zero
of potential energy is taken at the mountain top. What is her altitude?
Solution
If we measure the altitude from sea level,
U
(
y
)
!
U
(
1250 m
)
=
mg
(
y
!
1250 m
)
=
!
2
.
4
"
10
5
J
,
or
y
=
1250 m
!
2
.
4
!
10
5
J
=
(60 kg)
(
9
.
8 m/s
2
)
=
842 m
.
Problem 10 Solution.
Problem
17. A particle moves along the
x
-axis under the influence of a force
F
=
ax
2
+
b
,
where
a
and
b
are constants. Find its
potential energy as a function of position, taking
U
=
0
at
x
=
0.
Solution
Equation 8-2a, with
U
(0)
=
0,
gives
U
(
x
)
=
!
F
x
d
"
x
=
!
0
x
#
(
a
"
x
2
+
b
)
d
"
x
0
x
#
=
!
1
3
ax
3
!
bx
.

This
** preview**
has intentionally

**sections.**

*blurred***to view the full version.**

*Sign up*