Question

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

• A. $g(sin\theta +\sqrt{2}{\mu }_{k}cos\theta )$
• B. $g(sin\theta +{\mu }_{k}cos\theta )$
• 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 $g(sin\theta -\sqrt{2}{\mu }_{k}cos\theta )$

The force balance along the surface gives us the equation

$mgsin\theta -2{\mu }_{k}N=ma$ (there are 2 surfaces)

and perpendicular to it gives

$mgcos\theta =\sqrt{2}N$

Solving both the equations we get

$mgsin\theta -2{\mu }_k\frac{mgcos\theta }{\sqrt{2}}=ma$

or

$a=g(sin\theta -\sqrt{2}{\mu }_{k}cos\theta )$