A block is pushed onto a conveyor belt. The belt is moving at velocity $v_{_{0}} = 1 m / s,$ the block’s initial velocity $u_{0} = 2 m / s$ is perpendicular to the belt’s velocity. During its subsequent motion, what is the minimum velocity of the block with respect to the ground? The coefficient of friction is large enough to prevent the block from falling off the belt. If minimum velocity is $\frac{2}{\sqrt{n}} .$ Find $n$ =?

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answer is 5.

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Detailed Solution

w.r.t belt particle will move along straight line $v^{2} = {(1 - k)}^{2} + 4 k^{2}$

$\frac{d v^{2}}{d t} = - 2 (1 - k) + 8 k = 0$

$k = \frac{1}{5}$

$v^{2} = \frac{16}{25} + \frac{4}{25}$

$v = \frac{2}{\sqrt{5}}$

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