JEE MainPhysicsModerateNewton's Laws of Motion

Starting from rest, a body slides down a $45^{\circ}$ inclined
plane in twice the time it takes to slide down the same distance in the
absence of friction. The coefficient of friction between the body and
the inclined plane is

Show answer and solution

\mu = tan\theta\left( 1 - \frac{1}{n^{2}} \right)

\theta = 45^{\circ}

and

n = 2

(Given)

\therefore\mu = tan45^{\circ}\left( 1 - \frac{1}{2^{2}} \right) = 1 - \frac{1}{4} = \frac{3}{4} = 0.75

Source: Grass-G / P-06 Physics paper 14 October.docx

Need a video for this idea?

Scan any doubt from your notes and get a whiteboard explanation.