Question

A conveyor belt is moving at a constant speed of 2 m/s. A box is gently dropped on it. The coefficient of friction between them is $\mu =0.5$. The distance that the box will move relative to belt before coming to rest on it, taking g=10$m/s^2$ is

• A. 0.4m
• B. 1.2m
• C. 0.6m
• D. Zero

Answer 1

The correct option is A 0.4m

$F=\mu mg$
retardation of the block on the belt
$a=\frac{F}{m}=\mu g$
From $v^2=u^2+2as$
$0=2^2-2\left(\mu g\right)s$
$s=\frac{4}{2\times 0.5\times 10}=0.4m$