A conveyor belt is moving at a constant speed of $2\, ms^{-1}$. 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 ........ $m$.
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Force of friction, $f=\mu m g$
$\therefore a=\frac{\mathrm{f}}{\mathrm{m}}=\frac{\mu \mathrm{mg}}{\mathrm{m}}=\mu \mathrm{g}=0.5 \times 10=5 \mathrm{ms}^{-2}$
Using, $\mathrm{v}^{2}-\mathrm{u}^{2}=2 \mathrm{aS}$
${0^{2}-2^{2}=2(-5) \times 5} $
${\mathrm{S}=0.4 \mathrm{m}}$
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