In figure, the light spring is of force constant `k` and is on a smooth horizontal surface. Initially the spring is relaxed. Calculate the work done by an external agent to lower the hanging body of mass M slowly, till it remains in equilibrium.

Answer

`(m^2g^2)/(-2k)`

Explanation

Let the force applied by the man, the external agent, be F. The net force acting on the body is zero because it is lowered slowly.
Let the elongation of the spring at any instant be x.
Thus the spring pulls the body up by an amount `kx`.
`implies F_(n et)=F+kx-mg=0`
`impliesF=mg-kx` (i)
The work done `dW` by F in lowering the body through a distance `x_0` is
`W=int dW=underset0overset(x_0)int(mg-kx)dx`
`impliesW=mgx_0-1/2kx_0^2`

The body is lowered (slowly) till it remains in equilibrium. Therefore the external force F is variable and it is zero at the equilibrium of `m`, since the spring force `kx_0` counter balances the weight mg of the body
`implieskx_0=mgimpliesx_0=(mg)/(k)` (iii)
Putting `x_0=(mg)/(k)=` in Eqs. (ii), we obtain `W=(m^2g^2)/(-2k)`