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 $g=10{ms}^{-2}$is $:$

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

Answer 1

The correct option is A $0.4$ $m$

$a=\frac{F}{m}=\frac{\mu mg}{m}=\mu g$

From, $v^2=u^2+2$ as

$0=(2{)}^2-2(\mu g)s$

$s=\frac{4}{2\times 0.5\times 10}=0.4$m

Force,

$F=\mu g$

Retardation of the block in the belt

$a=\frac{F}{m}=\frac{\mu mg}{m}=\mu g$

From, $v^2=u^2+2$ as

$0=(2{)}^2-2(\mu g)s$

$s=\frac{4}{2\times 0.5\times 10}=0.4$m.