Chapter 8
Page 1
CHAPTER 8 - Conservation of Energy
1.
The potential energy of the spring is zero when the spring is not compressed (
x
= 0).
For the stored potential energy, we
have
U
=
!
kx
f
2
– 0;
35.0 J =
!
(82.0 N/m)
x
f
2
– 0, which gives
x
f
=
0.924 m
.
2.
For the potential energy change we have
∆
U
=
mg
∆
y
= (5.0 kg)(9.80 m/s
2
)(1.5 m) =
74 J
.
3.
For the potential energy change we have
∆
U
=
mg
∆
y
= (58 kg)(9.80 m/s
2
)(3.8 m) =
2.2
×
10
3
J
.
4.
(
a
) With the reference level at the ground, for the potential energy change we have
∆
U
=
mg
∆
y
= (66.5 kg)(9.80 m/s
2
)(2660 m – 1500 m) =
7.56
×
10
5
J
.
(
b
) The minimum work would be equal to the change in potential energy:
W
min
=
∆
U
=
7.56
×
10
5
J
.
(
c
)
Yes
,
the actual work will be more than this.
There will be additional work required for any
kinetic energy change, and to overcome retarding forces, such as air resistance and ground
deformation.
5.
(
a
) With the reference level at the ground, for the potential energy we have
U
a
=
mgy
a
= (2.20 kg)(9.80 m/s
2
)(2.40 m) =
51.7 J
.
(
b
) With the reference level at the top of the head, for the potential energy we have
U
b
=
mg
(
y
b
–
h
)= (2.20 kg)(9.80 m/s
2
)(2.40 m – 1.70 m) =
15.1 J
.
(
c
)
Because the person lifted the book from the reference level in part (
a
), the potential energy is equal
to the work done:
51.7 J
.
In part (
b
) the initial potential energy was negative, so the final
potential energy is not the work done, which was still 51.7 J.
6.
For the potential energy
U
= 3
x
2
+ 2
xy
+ 4
y
2
z
, we find the components of the force from
F
x
= –
∂
U
/
∂
x
= – 6
x
– 2
y
– 0 = – (6
x
+ 2
y
);
F
y
= –
∂
U
/
∂
y
= – 0 – 2
x
– 8
yz
= – (2
x
+ 8
yz
);
F
z
= –
∂
U
/
∂
z
= – 0 – 0 – 4
y
2
= – (4
y
2
).
Thus the force is
F
= – (6
x
+ 2
y
)
i
– (2
x
+ 8
yz
)
j
– (4
y
2
)
k
.
7.
(
a
) Because the force
F
= (–
kx
+
ax
3
+
bx
4
)
i
is a function only of position, it is
conservative
.
(
b
) We find the form of the potential energy function from
U
= –
∫
F
·
d
r
= –
∫
(–
kx
+
ax
3
+
bx
4
)
i
·
(d
x
i
+ d
y
j
+ d
z
k
)
= –
∫
(–
kx
+
ax
3
+
bx
4
)d
x
=
!
kx
2
–
#(
ax
4
–
$
bx
5
+ constant
.
8.
This force is
not conservative
.
If the object moves along a path that returns to the starting point, the
direction of the motion (the direction of the velocity) changes.
Because the direction of the force changes with the
change in direction of the motion, the net work done by the force is not zero.
9.
The potential energy of the spring is zero when the spring is not stretched or compressed (
x
= 0).
(
a
) For the change in potential energy, we have
∆
U
=
!
kx
2
–
!
kx
0
2
=
!
k
(
x
2
–
x
0
2
)
.
(
b

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