Question

A block lying on a long horizontal conveyor belt moving at a constant velocity, receives a velocity $6m/s$ relative to the ground in the direction opposite to the direction of motion of the conveyor. After $t=2s$, the velocity of the block becomes equal to the velocity of the belt. If the coefficient of friction between the block and the belt is $0.4$, then the velocity (in $m/s$) of the conveyor belt is

Answer 1

Given, coefficient of friction between the block and the belt, $\mu =0.4$
So, friction force on the block due to belt, $f=-\mu mg$
$a=f/m=-\mu g$
$=-0.4\times 10$
$=-4m/s^2$
Given, initial velocity of the block, $u=6m/s$
As given velocity of the block after $2s$ becomes equal to the velocity of the belt,
So, velocity of the belt,
$v=u+at$
$v=6-4\times 2$
$=-2m/s$