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KU
E
m
g
d
fk
=−
−
=
−
''
th
(sin
cos )
θ
μ
which leads to the speed at the bottom of the ramp
v
K
m
gd
f
k
==
−
=
2
25
5
ms
.
sin
cos
.
θμ
b
g
(b) This speed begins its horizontal motion, where
f
k
=
k
mg
and
Δ
U
= 0. It slides a
distance
d'
before it stops. According to Eq. 831 (with
W
= 0),
0
0
1
2
0
1
2
2
2
=++
++
′
−
+
′
ΔΔ
KUE
mv
mgd
gd
gd
k
kk
Δ
th
θ μ
sin
cos
b
g
ch
where we have divided by mass and substituted from part (a) in the last step. Therefore,
′ =
−
=
d
d
k
k
cos
..
bg
54m
(c) We see from the algebraic form of the results, above, that the answers do not depend
on mass. A 90 kg crate should have the same speed at the bottom and sliding distance
across the floor, to the extent that the friction relations in Ch. 6 are accurate. Interestingly,
since
g
does not appear in the relation for
d'
, the sliding distance would seem to be the
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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, Force, Work

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