A block is suspended by an ideal spring of force constant  $k.$  If the block is pulled down by applying constant force $F$  and if maximum displacement of block from its initial position of rest is $x_0$  then :

Initially, $kx=mg$...........(1)

here, x is the initial extension in spring.

Now after applying force F in downward direction, the spring will extend until the final velocity also becomes zero.

Using WET,

$W_g+W_F+W_S=K_f-K_i$

$\Rightarrow mg{x}_0+F{x}_0-\left[\frac{1}{2}k{\left(x+x_0\right)}^2-\frac{1}{2}k{x}^2\right]=0-0$...........(2)

Solving 1 and 2 we get,

$\boxed{x_0=\frac{2F}{K}}$

Work done by applied force $F$  is  $F{x}_0$.

Energy stored in the spring must be equal to work done by the force  $\frac{1}{2}K{x}_0^2=F{x}_0$