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Physics 221
Fall 2008
Homework #6 Solutions
Ch. 7
Due Tues, Oct 7, 2008
6.1
Consider a 0.142 kg baseball near the Earth’s surface.
The ball is thrown directly upward with an
initial speed
v
0
from an initial height of 2.0 m.
The ball reaches a height of 15.0 m before it
momentarily stops and then starts to fall back downward.
(a)
What is the change in the gravitational potential energy as the ball increases its height from 2.0
m to 15.0 m?
(The change in a quantity is the final value minus the initial value).
±
U
=
mg
(
y
2
–
y
1
)
=
(0.142 kg)(9.80 m/s
2
)(15.0 m – 2.0 m)
=
18.1 J .
(b)
Explicitly calculate how much work the gravitational force does on the ball during this height
change.
W
=
F
y
±
y
=
(–
mg
)(
y
2
–
y
1
)
=
–18.1 J .
(c)
In words, state how are the result in (b) is related to the result in (a).
The work done on an object by a conservative force is the negative of the change in potential
energy of the object associated with that force.
(d) Explain what the word “potential” in potential energy refers to.
That is, from part (c), the
potential energy can potentially do what?
Potential energy has the potential to do work.
(e)
Using the principle of conservation of mechanical energy, what is the initial speed
v
0
of the
ball?
The principle of conservation of mechanical energy states that the sum of the potential
U
and
kinetic
K
energies of an object is constant in time.
Alternatively, the sum of the changes in
these two quantities is zero, which gives
±
K
= –
±
U
.
Thus,
1
2
mv
2
2
±
1
2
mv
1
2
=
±²
U
.
Using part (a) gives
v
1
=
v
2
2
+
2
±
U
m
=
0
+
2(18.1 J)
0.142 kg
=
16.0 m/s
.
(f)
Explain what the term “conservative force” means, and then explain why the gravitational
force is a conservative force.
Explain why and how a potential energy can be associated with a
conservative force.
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 Fall '06
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