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KU
E
mv
mgd
i
ik
=+
ΔΔ
th
1
2
2
sin
cos
θμ
θ
bg
where
v
i
=
14
..
m s Dividing by mass and rearranging, we obtain
2
0.13m.
2 (sin
cos )
i
k
v
d
g
==
+
(b) Now that we know where on the incline it stops (
d'
= 0.13 + 0.55 = 0.68 m from the
bottom), we can use Eq. 833 again (with
W
= 0 and now with
K
i
= 0) to describe the
final kinetic energy (at the bottom):
E
mv
mgd
f
k
=−
−
=
′
−
th
1
2
2
cos
b
g
which — after dividing by the mass and rearranging — yields
vg
d
k
=
′
−=
22
7
cos
.
.
b
g
ms
(c) In part (a) it is clear that
d
increases if
μ
k
decreases — both mathematically (since it is
a positive term in the denominator) and intuitively (less friction — less energy “lost”). In
part (b), there are two terms in the expression for
v
which imply that it should increase if
k
were smaller: the increased value of
d'
=
d
0
+
d
and that last factor sin
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 Spring '08
 Any
 Physics, Energy, Force, Work

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