In the adjoining figure, block A is of mass m and block B is of mass 2 m  The spring has force constant k. All the surfaces are smooth and the system is released from rest with spring unstretched. 

 (a) Decrease in gravitational potential energy = increase in spring potential energy

$\therefore 2mg{x}_m=\frac{1}{2}k{x}_m^2$

$x_m=\frac{4mg}{k}$

(b)$E_i=E_f$

$0=\frac{1}{2}(m+2m)v^2+\frac{1}{2}\times k\times {\left(\frac{2mg}{k}\right)}^2-\left(2mg\right)\left(\frac{2mg}{k}\right)$

$v=2g\sqrt{\frac{m}{3k}}$

(c) $a=\frac{k{x}_m-2mg}{2m}$(upwards)

$=\frac{k\left(\frac{4mg}{k}\right)-2mg}{2m}=g$

(d) $T-2mg=ma$

$2mg-T=2ma$

Solving these two equations, we get 

$$\begin{gathered} a=0 \\ \text{ and } T=2mg \end{gathered}$$