A Block of mass m = 0.1kg  is connected to a spring of unknown spring constant k.  It is compressed to a distance x from its equilibrium position and released from rest.   After approaching the distance$\frac{x}{2}$   from equilibrium position , it hits another block and comes to rest momentarily, while the other block moves with a velocity 3m/s.   The total initial Energy of the spring is,

$$\begin{gathered} The block comes to rest after collision \\ \Rightarrow block velocity before collision was 3m/s, \\ before collision, KE of the spring block system is, \\ K.E=\frac{1}{2}m{v}^2 \\ substitute given values \\ \Rightarrow KE=\frac{1}{2}0.1\times 3^2 \\ \Rightarrow KE=0.45Joule---\left(1\right) \end{gathered}$$
   $$\begin{gathered} P.E.=\frac{1}{2}{\text{kx}}^2 \\ at displacement=\frac{x}{2}, \\ PE\text{=\hspace{0.17em}}\frac{1}{2}\text{k(}\frac{\text{x}}{2}{\text{)}}^2\text{---(2)} \\ Total energy =\frac{1}{2}{\text{kA}}^2 \\ at displacement x, \\ TE=\frac{1}{2}{\text{kx}}^2\text{---(3)} \\ substitute equation\left(3\right) in equation \left(2\right) \\ PE=\frac{1}{4}TE then remaining energy will be KE \\ KE=\frac{3}{4}TE \\ substitute equation\left(1\right) \\ 0.45 =\frac{3}{4}TE \\ TE=0.45 X\frac{4}{3} \\ Total energy =0.6 Joule \end{gathered}$$
 $\therefore T.E=\frac{4}{3}\times 0.45=0.6J$