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Question

Block of mass m is released from rest when spring is in its natural length. Assume pulley is ideal and block does not strike on ground during its motion in vertical plane. Then:

A
Maximum elongation in spring is 4mgk
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B
Maximum elongation in spring is 2mgk
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C
Maximum speed of block is 2gmk
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D
Maximum speed of block is gmk
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Solution

The correct option is C Maximum speed of block is 2g√mkWhen block is released, the spring will stretch by x and block will move downwards by 2x. By energy conservation principle, Energy gain by spring = Loss in gravitational potential energy by block 12kx2max=mg(2xmax) ⇒xmax=4mgk ( at maximum elongation) The block will attain maximum velocity when net force on the pulley is zero. Let net force on block be zero when x=x′ At equilibrium, kx′=2mg ⇒x′=2mgk The kinetic energy of the block is given by 12mv2=mg(2x)−12kx2 ∴12mv2max=mg(2x′)−12kx′2 Substituting x′ to get maximum velocity (vmax) 12mv2max=2mg(2mgk)−12k(2mgk)2 ⇒12mv2max=2m2g2k ⇒vmax=2g√mk Hence, options (a) and (c) are the correct answers.

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