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A block of mass m, lying on a smooth horizontal surface, is attached to a spring (of negligible mass) of spring constant k. The other end of the spring is fixed, as shown in the figure. The block is initally at rest in its equilibrium position. If now the block is pulled with a constant force F, the maximum speed of the block is :

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$\frac{2 F}{\sqrt{mk}}$

$\frac{F}{\sqrt{mk}}$

$\frac{F}{π \sqrt{mk}}$

$\frac{πF}{\sqrt{mk}}$

answer is C.

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Detailed Solution

At mean position, maximum speed occurs (equilibrium),

$F = k x$

$x = \frac{F}{k}$

$W_{F} + W_{s p} = ∆ K E$

$F (x) - \frac{1}{2} k {(\frac{F}{k})}^{2} = \frac{1}{2} m v^{2}$

$v_{m a x} = \frac{F}{\sqrt{m k}}$

Hence the correct answer is $\frac{F}{\sqrt{m k}} .$

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