p08_010.pdf

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10. We use Eq. 8-17, representing the conservation of mechanical energy (which neglects friction and other
dissipative effects).
(a) In the solution to exercise 2 (to which this problem refers), we found
U
i
=
mgy
i
= 196 J and
U
f
=
mgy
f
= 29 J (assuming the reference position is at the ground). Since
K
i
= 0 in this case,
we have
K
i
+
U
i
=
0 + 196 =
which gives
K
f
= 167 J and thus leads to
v
=
2K
f
=
m
2(167)
= 12.9 m/s
.
2.00
K
f
+
U
f
K
f
+ 29
(b) If we proceed algebraically through the calculation in part (a), we find
K
f
=
−∆U
=
mgh
where
h
=
y
i
−
y
f
and is positive-valued. Thus,
v
=
2K
f
=
m
2gh
as we might also have derived from the equations of Table 2-1 (particularly Eq. 2-16). The fact
that the answer is independent of mass means that the answer to part (b) is identical to that of
part (a).
(c) If
K
i
= 0, then we find
K
f
=
mgh
+
K
i
(where
K
i
is necessarily positive-valued). This represents
a larger value for
K
f
than in the previous parts, and thus leads to a larger value for
v.

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