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# Work, Kinetic Energy and Potential Energy

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146
CHAPTER 6. WORK, KINETIC ENERGY AND POTENTIAL ENERGY
R
mg
N
v
Dir. of
accel.
Figure 6.11:
Forces acting on the bead when it is at point A (the top of the loop).
v
2
= 2g(h - 2R) = 2g(3.50R - 2R) = 2g(1.5R) = 3.0 gR
(6.28)
and finally
v =
3.0 gR .
Since we don't have a numerical value for R, that's as far as we can go.
In the next part of the problem, we think about the forces acting on the bead at point A.
These are diagrammed in Fig. 6.11. Gravity pulls down on the bead with a force mg. There
is also a normal force from the track which I have drawn as having a downward component
N. But it is possible for the track to be pushing upward on the bead; if we get a negative
value for N we'll know that the track was pushing up.
At the top of the track the bead is moving on a circular path of radius R, with speed v.
So it is accelerating toward the center of the circle, namely downward. We know that the
downward forces must add up to give the centripetal force mv
2
/R:
mg + N =
mv
2
R
=
n =
mv
2
R
- mg = m
v
2
R
- g
.
But we can use our result from Eq. 6.28 to substitute for v
2
. This gives:
N = m
3.0 gR
R
- g = m(2g) = 2mg
Plug in the numbers:
N = 2(5.00 × 10
-3
kg)(9.80
m
s
2
) = 9.80 × 10
-2
N
At point A the track is pushing downward with a force of 9.80 × 10
-2
N.
14. Two children are playing a game in which they try to hit a small box on the
floor with a marble fired from a spring ­loaded gun that is mounted on the table.
The target box is 2.20 m horizontally from the edge of the table; see Fig. 6.12.

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