Question

A body of mass m is attached to the spring which is elongated to 25 cm by an applied force from it's equilibrium position the potential energy stored in the spring mass system is

Answer 1

Actually both the answers are correct. If you want to find the extension in spring when the block is in equilibrium then you should write an equation making net force on the block equal to zero.

So in equilibrium:

Kx=mg => x=mg/k.

But if you drop the block when the spring is in its natural length then block will go upto x=2mg/k because when it will pass through its equilibrium stage its acceleration would be zero but its velocity will not be zero hence it will go down further until its velocity becomes zero.

Hence by applying work energy theorem:

Mgx=(1/2)kx^2 => x=2Mg/k.

So if you will let the block come down slowly then elongation of spring will be Mg/k but if you will just drop the block after attaching it to spring then elongation will be 2Mg/k.

Hope you understand.