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Block A of mass M in the figure is released from rest when the extension in the spring is $x_{0}$ . The maximum downward displacement of the block is (assume $x_{0} < \frac{Mg}{k}$ )

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2(Mgk-x0)

Mg2k+x0

2Mgk-x0

2Mgk+x0

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detailed solution

Correct option is A

Loss in gravitational potential energy of block is equal to gain in elastic potential energy of springMgx = 12k(x+x0)2-12kx02On solving x = 2(Mgk-x0)

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Block A of mass M in the figure is released from rest when the extension in the spring is x0. The maximum downward displacement of the block is (assume x0< Mgk)

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Ready to Test Your Skills?

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Block A of mass M in the figure is released from rest when the extension in the spring is $x_{0}$ . The maximum downward displacement of the block is (assume $x_{0} < \frac{Mg}{k}$ )