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51. (a) The initial potential energy is
Um
g
y
ii
==
=
×
(520
1
kg) 9.8m s (300 m)
.53 10 J
2
6
di
where +
y
is upward and
y
= 0 at the bottom (so that
U
f
= 0).
(b) Since
f
k
=
μ
k
F
N
=
k
mg
cos
θ
we have
th
cos
kk
Ef
d
m
g
d
Δ=
=
from Eq. 831.
Now, the hillside surface (of length
d
= 500 m) is treated as an hypotenuse of a 345
triangle, so cos
=
x
/
d
where
x
= 400 m. Therefore,
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 Spring '08
 Any
 Physics, Energy, Potential Energy

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