A cubic block of mass m is sliding down on an inclined plane at 60° with an acceleration of g/2 , ​

Question

A cubic block of mass m is sliding down on an inclined plane at \(60 ^\circ\) with an acceleration of \(\frac{g}{2}\), the value of coefficient of kinetic friction is

(1) \(\sqrt 3 -1 \)

(2) \(\frac{\sqrt 3}{2}\)

(3) \(\frac{\sqrt2}{3}\)

(4) \(1- \frac{\sqrt3}{2}\)

Answer 1

Correct option is : (1) \(\sqrt 3 -1\) 

mgsin\(60 ^\circ\) – \(\mu\)mg\(\cos 60^\circ\) = ma

\(g \sin 60 - \mu g \cos 60 = \frac{g}{2}\)

\(\frac{\sqrt 3}{2} - \frac{\mu}{2} = \frac{1}{2}\)

\(\mu = \sqrt 3 -1\)