Answer:
As the block is sliding down the
inclined plane, force of friction F acts up the plane, Fig. 3(b).38. If a is
acceleration produced in the block, then net force on the block down the plane
is \[f=ma=mg\,\sin \text{ }\!\!\theta\!\!\text{ }\,\text{-
F}\,\text{=}\,mg\,\sin \text{ }\!\!\theta\!\!\text{ }-\mu mg\,\cos \text{
}\!\!\theta\!\!\text{ }=mg\left( \sin \text{ }\!\!\theta\!\!\text{ }-\mu \cos
\text{ }\!\!\theta\!\!\text{ } \right)\]
or \[a=g\left( \sin \theta -\mu \cos \theta \right)=10\left( \sin
{{30}^{\text{o}}}-0.2\cos {{30}^{\text{o}}} \right)=10\left(
\frac{1}{2}-0.2\frac{\sqrt{3}}{2} \right)\]
\[a=5\left( 1-0.2\times 1.732 \right)=5\times 0.6536=3.268\,m/{{s}^{2}}\]
Now, \[s=6.4\,m\]; u = 0 as the block starts
from rest.
t = ?
From \[s=ut=\frac{1}{2}a{{t}^{2}}\]
\[6.4=0+\frac{1}{2}\times 3.268{{t}^{2}}\]
\[t=\sqrt{\frac{2\times
6.4}{3.268}}=1.98s\]
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