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# Work, Kinetic Energy and Potential Energy

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6.2. WORKED EXAMPLES
145
R
A
h
Figure 6.10:
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?
[Ser4 8-11]
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
i
= 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
U
i
= mgh
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
f
=
1
2
mv
2
. At position A its height is 2R
(it is a full diameter above the "ground level" of the track) so its potential energy is
U
f
= mg(2R) = 2mgR .
The total energy of the bead is conserved: K
i
+ U
i
= K
f
+ U
f
. This gives us:
0 + mgh =
1
2
mv
2
+ 2mgR ,
where we want to solve for v (the speed at A). The mass m cancels out, giving:
gh =
1
2
v
2
= 2gR
=
1
2
v
2
= gh - 2gR = g(h - 2R)
and then

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Summary :

## 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

Tags : energy,track,potential,speed,height,mass,kinetic,force,position,then,but,mgh,released
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