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A ball of mass m is attached to the lower end of a light vertical spring of force constant $k$ . The upper end of the spring is fixed. The ball is released from rest with the spring at its normal (unstretched) length, and comes to rest again after descending through a distance $x$ :

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$x = \frac{m g}{k}$

$x = \frac{2 m g}{k}$

the ball will have no acceleration at the position where it has descended through $\frac{x}{2}$

the ball will have an upward acceleration equal to $g$ at its lowermost position.

answer is B, C, D.

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Detailed Solution

$\frac{1}{2} k x^{2} = m g x, ∴ x = \frac{2 m g}{k}$

At $\frac{x}{2},$ force in the spring, $F = k \frac{x}{2} = k \frac{m g}{2} = m g,$ so, net force on the block is zero. Also at extreme position the block has net upward force and so net upward acceleration.

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