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The figure shows a mass m on a frictionless surface. It is connected to rigid wall by the mean of a massless spring of its constant k. Initially, the spring is at its natural position. If a force of constant magnitude starts acting on the block towards right, then the speed of the block when the deformation in spring is x, will be [AIIMS 2017]

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$\sqrt{\frac{F \cdot x - k x^{2}}{m}}$

$\sqrt{\frac{x (F - k)}{m}}$

$\sqrt{\frac{2 F \cdot x - k x^{2}}{m}}$

$\sqrt{\frac{F \cdot x - k x^{2}}{2 m}}$

answer is A.

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Detailed Solution

Free body diagram of block is as shown below.

Now, from law of conservation of energy, $W = Δ K$

$\Rightarrow W_{F} + W_{s p} = \frac{1}{2} m v^{2}$

$\Rightarrow F \cdot x - \frac{1}{2} k x^{2} = \frac{1}{2} m v^{2}$

$∴ v = \sqrt{\frac{2 F \cdot x - k x^{2}}{m}}$

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