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top
ΔΔ
KU
KK
m
g
L
+=
−+
=
0
0
0
which, upon requiring
K
top
= 0, gives
K
0
=
mgL
and thus leads to
2
0
0
2
2
2(9.80 m/s )(0.452 m)
2.98 m/s
K
vg
L
m
==
=
=
.
(b) We also found in the Problem 4 that the potential energy change is
Δ
U
= –
mgL
in
going from the initial point to the lowest point (the bottom). Thus,
bottom
K
U
m
g
L
−−
=
0
0
0
which, with
K
0
=
mgL
, leads to
K
bottom
= 2
mgL
. Therefore,
2
bottom
bottom
2
4
4(9.80 m/s )(0.452 m)
4.21 m/s
K
L
m
=
=
.
(c) Since there is no change in height (going from initial point to the rightmost point),
then
Δ
U
= 0, which implies
Δ
K
= 0. Consequently, the speed is the same as what it was
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 Spring '08
 Any
 Physics, Energy, Friction

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