A block takes $t$ time to slide down a plane inclined at $45^\circ$ to the horizontal. If the surface is made smooth (frictionless), the block takes time $\dfrac{t}{2}$ to slide down the plane. The coefficient of friction between the block and the inclined plane is $\left(\dfrac{\alpha}{100}\right)$. The value of $\alpha$ is ______.
Numerical value type. Enter your answer.
Answer: 75
Concept. Same length from rest: $L=\tfrac12at^2$, so $t\propto\dfrac1{\sqrt a}$. Smooth: $a_s=g\sin\theta$; rough: $a_r=g(\sin\theta-\mu\cos\theta)$.
Step 1. Ratio of accelerations. $t_s=\dfrac t2\Rightarrow\dfrac{a_s}{a_r}=\left(\dfrac{t}{t/2}\right)^2=4$.
Step 2. At $45^\circ$ ($\sin=\cos$): $\dfrac{1}{1-\mu}=4\Rightarrow\mu=0.75$.
Answer. $\alpha=75$.
Solution by Sreeraj P, M.Sc Physics
