JEE Main · 2023 · 25 Jan 2023 · Shift 2 · Laws of Motion

On the $45^{\circ }$ incline the push needed to start the block moving up must overcome gravity along the slope and friction together, $F_1=mg\sin 45^{\circ }+\mu mg\cos 45^{\circ }$, while the push needed to just stop it sliding down is the shortfall of gravity over friction, $F_2=mg\sin 45^{\circ }-\mu mg\cos 45^{\circ }$. Setting $F_1=2F_2$ and cancelling $\frac{mg}{\sqrt{2}}$ gives $1+\mu =2(1-\mu )$, hence $3\mu =1$ and the coefficient of friction is $\mu =\frac{1}{3}\approx 0.33$.