2
2(167 J)
12.9 m/s.
2.00 kg
f
K
v
m
==
=
(b) If we proceed algebraically through the calculation in part (a), we find
K
f
= –
Δ
U
=
mgh
where
h
=
y
i
–
y
f
and is positivevalued. Thus,
2
2
f
K
vg
h
m
as we might also have derived from the equations of Table 21 (particularly Eq. 216).
The fact that the answer is independent of mass means that the answer to part (b) is
identical to that of part (a), i.e.,
12.9 m/s
v
=
.
(c) If
K
i
≠
0
, then we find
K
f
=
mgh
+
K
i
(where
K
i
is necessarily positivevalued). This
represents a larger value for
K
f
than in the previous parts, and thus leads to a larger value
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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, Friction

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