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

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Aicharya · Accepted Answer

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, $\begin{aligned} v^2 & =u^2+2 a s \ 0 & =(2)^2-2(\mu g) s \ s & =\frac{4}{2 	imes 0.5 	imes 10}=0.4 \mathrm{~m} \end{aligned}$

A conveyor belt is moving at a constant speed of $2 \mathrm{~m} / \mathrm{s}$. A box is gently dropped on it…

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