Question

A heavy block of mass $M$ is slowly placed on a conveyer belt moving with a speed $v$. The coefficient of friction between the block and the belt is $\mu$. Through what distance will the block slide on the belt ?

• A. $\frac{v}{\mu g}$
• B. $\frac{v^2}{\mu g}$
• C. $\frac{v}{2\mu g}$
• D. $\frac{v^2}{2\mu g}$

Answer 1

The correct option is D $\frac{v^2}{2\mu g}$

When the block is placed on the belt, force of kinetic friction acts on the block until it attains the same speed as that of belt.

Frictional force acting on the block = $\mu mg$

This force causes the acceleration of the block.

$⟹v\frac{dv}{dx}=\mu g$

$⟹{\int }_0^{v}vdv=\mu g{\int }_0^{s}dx$

$⟹\frac{v^2}{2}=\mu gs$

$⟹s=\frac{v^2}{2\mu g}$