Physics - NLM Question with Solution | TestHub

At equilibrium the spring is stretched by $mg/k$. Natural length is $mg/k$ above the equilibrium position; pushing up by $mg/2k$ leaves the spring stretched by $mg/2k$.

Spring force (up) $=k\cdot \frac{mg}{2k}=\frac{mg}{2}$. Net downward force $=mg-\frac{mg}{2}=\frac{mg}{2}$, so $a=\frac{g}{2}$.