Question

An object is gently placed on a long conveyer belt moving with a uniform speed of $5m{s}^{-1}$. If the coefficient of friction between the block and the belt is 0.5, how far will the box move on the belt (relative to belt) before coming to rest? Take $g=10m{s}^{-2}$.

• A. 2.0 m
• B. 2.5 m
• C. 3.0 m
• D. 3.5 m

Answer 1

The correct option is B

2.5 m

The force of friction between the block and the belt is $f=\mu mg,$ where m is the mass of the object. This force produces an acceleration of the block which is given by
$a=\frac{f}{m}=\frac{\mu mg}{m}=\mu g$
The block will move relative to the belt until its speed (v) becomes equal to the speed of the belt. Since u = 0, we have
$v^2=u^2+2as=2as\Rightarrow s=\frac{v^2}{2a}=\frac{v^2}{2\mu g}=\frac{(5{)}^2}{2\times 0.5\times 10}=2.5m$
Hence, the correct choice is (b).