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Q.

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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a

The loss of gravitational potential energy of the ball is equal to twice the gain in kinetic energy for the time interval when it has descended through x/2.

b

The ball will have equal kinetic energy and spring energy at the moment when it has descended through  x/2

c

The ball will have an upward acceleration equal in magnitude to its initial acceleration, when it is at its lowermost position. 

d

The ball will have no acceleration at the position where it has descended through x/2.

answer is A, B, C, D.

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Detailed Solution

12kx2=mgx

x=2mgk

at x=mgk

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kmgkmg=ma

a = 0

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at x=2mgk 

2mg – mg = ma

a=g

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