In Case 1, a mass M hangs from a vertical spring having spring constant k and is at rest in its equilibrium position. In Case 2 the mass has been lifted a distance D vertically upward. If we define the potential energy in Case 1 to be zero, what is the potential energy of Case 2

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answer :  1/2 KD^2  ( option A )

Explanation:

P.E ( potential energy ) = mgd

In case 1 P.E = 0   i.e. mgd = 0  

Given that in case 2 the Mass M had moved through the Distance D by the compression of the spring

The potential energy of the M in case 2

= P.E of M at rest + P.E of the spring

= 0 + 1/2 KD^2