A conveyor belt is moving at a constant speed of $2 \mathrm{~m} / \mathrm{s}$. 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 $g=10 \mathrm{~ms}^{-2}$, is

• A

  $1.2 \mathrm{~m}$
• B

  $0.6 \mathrm{~m}$
• C

  zero
• Correct answer
  D

  $0.4 \mathrm{~m}$

Solution

Force, $\quad F=\mu m g$ Retardation of the block on the belt $$ a=\frac{F}{m}=\frac{\mu m g}{m}=\mu g $$ From, $$ \begin{aligned} v^2 & =u^2+2 a s \\ 0 & =(2)^2-2(\mu g) s \\ s & =\frac{4}{2 \times 0.5 \times 10}=0.4 \mathrm{~m} \end{aligned} $$

Hence, the correct answer is option: 'D'.