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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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Verified Answer
The correct answer is:
$0.4 \mathrm{~m}$
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,
$$
\begin{aligned}
v^2 & =u^2+2 a s \\
0 & =(2)^2-2(\mu g) s \\
s & =\frac{4}{2 \times 0.5 \times 10}=0.4 \mathrm{~m}
\end{aligned}
$$
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