Work, Kinetic Energy and Potential Energy
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6.2. WORKED EXAMPLES
Bead slides on track in Example 13.
13. A bead slides without friction on a looptheloop track (see Fig. 6.10). If
the bead is released from a height h = 3.50R, what is its speed at point A? How
large is the normal force on it if its mass is 5.00 g?
In this problem, there are no friction forces acting on the particle (the bead). Gravity acts
on it and gravity is a conservative force. The track will exert a normal forces on the bead,
but this force does no work. So the total energy of the bead --kinetic plus (gravitational)
potential energy-- will be conserved.
At the initial position, when the bead is released, the bead has no speed; K
= 0. But
it is at a height h above the bottom of the track. If we agree to measure height from the
bottom of the track, then the initial potential energy of the bead is
where m = 5.00 g is the mass of the bead.
At the final position (A), the bead has both kinetic and potential energy. If the bead's
speed at A is v, then its final kinetic energy is K
. At position A its height is 2R
(it is a full diameter above the "ground level" of the track) so its potential energy is
= mg(2R) = 2mgR .
The total energy of the bead is conserved: K
. This gives us:
0 + mgh =
+ 2mgR ,
where we want to solve for v (the speed at A). The mass m cancels out, giving:
= gh - 2gR = g(h - 2R)
If we agree to measure height from the bottom of the track, then the initial potential energy of the bead is U i = mgh where m = 5.00 g is the mass of the bead. The mass m cancels out, giving: gh = 1 2 v 2 = 2gR = 1 2 v 2 = gh - 2gR = g(h - 2R) and then