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A block of mass $m$ moves on a horizontal circle against the wall of a cylindrical room of radius $R$ . The floor of the room, on which the block moves, is smooth but the friction coefficient between the wall and the block is $μ$ . The block is given an initial speed $V_{0}$ . The power developed by the resultant force acting on the block as a function of distance travelled $s$ is

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$\frac{μ m_{0}^{3}}{R} e^{\frac{- 3 s}{μ}}$

$\frac{μ m V_{0}^{3}}{R} e^{\frac{- 3 μ s}{R}}$

$\frac{μ m V_{0}^{3}}{R}$

answer is B.

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Detailed Solution

$f = - m a$

$\Rightarrow μ N = - m a$

$\Rightarrow μ \frac{m V^{2}}{R} = - m \times V \frac{d V}{d S}$

$\Rightarrow \frac{μ V}{R} = \frac{- d V}{d S}$

$\Rightarrow \frac{d V}{V} = \frac{- μ}{R} d S$

$\Rightarrow \int_{v_{0}}^{v} \frac{d V}{V} = \frac{- μ}{R} \int_{0}^{s} d S$

$\Rightarrow I n |\frac{V}{V_{0}}| = \frac{- μ}{R} (S)$

$\Rightarrow V = V_{0} e^{\frac{- μ S}{R}}$ -------------(1)

Now power consumed by friction,

$P = - f V$

$= - μ N V$

$= - μ \frac{m V^{2}}{R} V$

$= - μ \frac{m V^{3}}{R}$ --------------(2)

substitute the value of V from eq(1), we get

$P = \frac{- μ m}{R} V_{0}^{3} e^{\frac{- 3 μ S}{R}}$

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