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# Work, Kinetic Energy and Potential Energy

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6.2. WORKED EXAMPLES
143
60
o
2.0 kg
4.0 m
8.0 m/s
(a)
(b)
v= ?
Figure 6.8:
(a) Pendulum in Example 12 swings through lowest point. (b) Pendulum has swung 60
past
lowest point.
bottom of its swing. Then at the bottom of the swing the stone has zero potential energy,
while its kinetic energy is
K
i
=
1
2
mv
2
0
=
1
2
(2.0 kg)(8.0
m
s
)
2
= 64 J
When the stone has swung up by 60
(as in Fig. 6.8(b)) it has some potential energy. To
figure out how much, we need to calculate the height of the stone above the lowest point of
the swing
. By simple geometry, the stone's position is
(4.0 m) cos 60
= 2.0 m
down from the top of the string, so it must be
4.0 m - 2.0 m = 2.0 m
up from the lowest point. So its potential energy at this point is
U
f
= mgy = (2.0 kg)(9.80
m
s
2
)(2.0 m) = 39.2 J
It will also have a kinetic energy K
f
=
1
2
mv
2
f
, where v
f
is the final speed.
Now in this system there are only a conservative force acting on the particle of interest,
i.e. the stone. (We should note that the string tension also acts on the stone, but since
it always pulls perpendicularly to the motion of the stone, it does no work.) So the total
mechanical energy of the stone is conserved:
K
i
+ U
i
= K
f
+ U
f
We can substitute the values found above to get:
64.0 J + 0 =
1
2
(2.0 kg)v
2
f
+ 39.2 J
which we can solve for v
f
:
(1.0 kg)v
2
f
= 64.0 J - 39.2 J = 24.8 J
=
v
2
f
= 24.8
m
2
s
2

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Summary :

## (We should note that the string tension also acts on the stone, but since it always pulls perpendicularly to the motion of the stone, it does no work.) So the total mechanical energy of the stone is conserved: K i + U i = K f + U f We can substitute the values found above to get: 64.0 J + 0 = 1 2 (2.0 kg)v 2 f + 39.2 J which we can solve for v f : (1.0 kg)v 2 f = 64.0 J - 39.2 J = 24.8 J = v 2 f = 24.8 m 2 s 2

Tags : stone,energy,point,lowest,swing,392,potential,248,640,pendulum,swung,aboe,bottom
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