question_answer

In figure, a crate slides down an inclined right angled through. The coefficient of kinetic friction between the crate and the trough is \[{{\mu }_{k}}.\] The acceleration of the crate is:

A) \[\operatorname{g}\,sin\theta \]

B) \[\operatorname{g}\,sin\theta -{{\mu }_{k}}g\cos \theta \]

C) \[\operatorname{g}\,(sin\theta -\sqrt{2}{{\mu }_{k}}\cos \theta )\]

D) None of these

Correct Answer: C

Solution :

If \[N\]is the normal reaction of each block, then
\[2N\cos 45{}^\circ =mg\cos \theta \]
\[\therefore \]\[N=\frac{mg\cos \theta }{\sqrt{2}}\]
Frictional force on the block
\[f-{{\mu }_{\operatorname{k}}}\left( 2N \right)\]
\[=\sqrt{2}{{\mu }_{k}}\operatorname{mg}\cos \theta \]	...(i)
Now \[mg\sin \theta -f=ma\]	...(ii)
From above equations, we get
\[a=g\left( \sin \theta -\sqrt{2}{{\mu }_{k}}\cos \theta  \right).\]