System shown in figure is released from rest with the spring at natural length. Pulley and spring is massless and friction is absent everywhere. The speed (in m/s) of 5kg  block when 2kg  block leaves the contact with ground is (Take force constant of spring $k=40\mathrm{N}/\mathrm{m}\text{ and }g=10\mathrm{m}/{\mathrm{s}}^2,\sqrt{2}=1.414$)

2kg block will leave the contact if   $T=mg \left(here m=2\text{\hspace{0.17em}kg}\right)$
Also,  $T=k{x}_0$

$\Rightarrow x_0=\frac{1}{2}m$
    From energy conservation for system 
Decrease in gravitational potential energy = Increase in spring potential energy + Increase in kinetic energy
$mgx=\frac{1}{2}k{x}^2+\frac{1}{2}m{v}^2$
where  $m=5\text{\hspace{0.17em}kg},x=\frac{1}{2}\text{meter},k=40\text{N}/\text{m}$
$\begin{array}{l}\Rightarrow 5\times 10\times \frac{1}{2}=\frac{1}{2}\times 40\times {\left(\frac{1}{2}\right)}^2+\frac{1}{2}\times 5v^2\\ \Rightarrow 40=5v^2\\ \Rightarrow v^2=8\\ \Rightarrow v=2\sqrt{2}\end{array}$
Therefore, the correct answer is 2.83