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With
μ
k
= 0.25, Eq. 62 leads to
f
k
= 11 N.
(b) We apply Newton’s second law to the
x
axis:
()
2
15 N cos 40
11 N
cos
0.14 m/s
3.5 kg
k
Ff
m
a
a
θ
°−
−=
¡
==
.
Since the result is positivevalued, then the block is accelerating in the +
x
(rightward)
direction.
7. We choose +
x
horizontally rightwards and +
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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, Force

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