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9. We neglect any work done by friction. We work with SI units, so the speed is
converted:
v
= 130(1000/3600) = 36.1 m/s.
(a) We use Eq. 817:
K
f
+
U
f
=
K
i
+
U
i
with
U
i
= 0,
U
f
=
mgh
and
K
f
= 0. Since
Km
v
i
=
1
2
2
, where
v
is the initial speed of the truck, we obtain
22
2
2
1
(36.1m/s)
66.5 m
2
2
2(9.8 m/s )
v
mv
mgh
h
g
=
¡
==
=
.
If
L
is the length of the ramp, then
L
sin 15° = 66.5 m so that
L
= (66.5 m)/sin 15° = 257
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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, Friction, Work

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