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A conveyor belt is moving at a constant speed of $2 m / s .$ A box is gently dropped on it. The coefficient of friction between them is $µ = 0.5 .$ The distance that the box will move relative to belt before coming to rest on it taking $g = 10 {ms}^{- 2},$ is

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$1.2 m$

$0.6 m$

zero

$0.4 m$

answer is D.

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Detailed Solution

Frictional force on the box $f = µ mg$

There will be relative motion till the velocities are different.

$∴$ Acceleration in the box

$\begin{array}{l} a = μ g = 5 {ms}^{- 2} \\ v^{2} = u^{2} + 2 as \\ \Rightarrow 0 = 2^{2} + 2 \times (5) s \\ \Rightarrow s = - \frac{2}{5} w.r.t. belt \\ \Rightarrow distance = 0.4 m \end{array}$

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