Note: one might wish to check that the skier stays in contact with the hill — which is
indeed the case, here. For instance, at
A
we find
v
2
/
r
≈
2 m/s
2
which is considerably less
than
g
.
(b) With
K
A
= 0, we have
KU
K
m
g
y
BBAA
B
A
+=+
¡
00
which yields
K
B
= 724 J, and the corresponding speed is
vK
m
==
24
9
.m
s
.
(c) Expressed in terms of mass, we have
mv
mgy
mv
mgy
BB
AA
¡
+=
+
1
2
1
2
22
.
Thus, the mass
m
cancels, and we observe that solving for speed does not depend on the
value of mass (or weight).
110. We take her original elevation to be the
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This note was uploaded on 05/17/2011 for the course PHY 2049 taught by Professor Any during the Spring '08 term at University of Florida.
 Spring '08
 Any
 Physics, Energy, Mass

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