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In the adjoining diagram, the ball A is released from rest when the spring is at its natural length (neither stretched nor compressed). For the block B of mass 20 kg to leave contact with the ground the minimum mass of A in kg must be

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answer is 10.

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Detailed Solution

If x be the required extension in the spring, then
kx = Mg $∴ x = \frac{M g}{k}$
now for ball A,
$m g x = \frac{1}{2} k x^{2} o r m g = \frac{k}{2} (\frac{M g}{k}) = \frac{M}{2} = \frac{20}{2} = 10 k g$

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