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2
12
sin
k
mg
f
a
mm
θ
−
=
+
For
f
k
=
μ
k
F
N
=
k
m
1
g
,
we obtain
22
2
(3.0 kg)(9.8 m/s )sin30
(0.25)(2.0 kg)(9.8 m/s )
1.96 m/s
3.0 kg 2.0 kg
a
°−
==
+
.
Returning this value to either of the above two equations, we find
T
= 8.8 N.
64. Note that since no static friction coefficient is mentioned, we assume
f
s
is not relevant
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 Spring '08
 Any
 Physics, Friction

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