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In the position shown in figure, the spring is at its natural length. The block of mass m is given a velocity v₀ towards the vertical support at t = 0. The coefficient of friction between the block and the surface is given by $μ = αx$ , where $α$ a positive constant and x is the position of the block from its starting position. The block comes to rest for the first time at x, which is

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$v_{0} \sqrt{\frac{m}{k + αmg}}$

$v_{0} \sqrt{\frac{m}{k}}$

$v_{0} \sqrt{\frac{m}{αg}}$

None of these

answer is A.

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Detailed Solution

According to the work-energy theorem,

$\frac{1}{2} {mv}_{0}^{2} = mgα \int_{0}^{x} xdx + \frac{1}{2} {kx}^{2}$

Solving we get, $x = v_{0} \sqrt{\frac{m}{k + αmg}}$

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