Question

A block of mass $m$ slides down an inclined right angled trough. If the coefficient of kinetic friction between the block and the trough is ${\mu }_k$, acceleration of the block down the plane is:

• A. $g(\sin \theta -2{\mu }_k\cos \theta )$
• B. $g(\sin \theta +2{\mu }_k\cos \theta )$
• C. $g(\sin \theta -\sqrt{2}{\mu }_k\cos \theta )$
• D. $g(\sin \theta -{\mu }_k\cos \theta )$

Answer 1

The correct option is C $g(\sin \theta -\sqrt{2}{\mu }_k\cos \theta )$

Here the frictional forces are perpendicular to the plane. As there are two surfaces,

So, frictional force ,$f=2{\mu }_{k}N$
Here also $mg$ downward from the plane. thus,
$mg\cos \theta -N\sqrt{2}=0....\left(1\right)$
$ma=mg\sin \theta -f=mg\sin \theta -2{\mu }_{k}N...\left(2\right)$
using (1), from (1), $ma=mg\sin \theta -2{\mu }_k\frac{mg\cos \theta }{\sqrt{2}}$
or, $a=g(\sin \theta -\sqrt{2}{\mu }_k\cos \theta )$