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 $\text{m}/\text{s}$) of $5\text{kg}$ block when $2\text{kg}$ block leaves the contact with ground is (Take force constant of spring $K=40\text{N}/\text{m}$ and $g=10\text{m}/{\text{s}}^{\text{2}},\sqrt{2}=1.414$)

1.

 $2\text{kg}$block will leave the contact if $T=mg$$\left(m=2\text{kg}\right)$

Also, $T=k{x}_0\Rightarrow x_0=\frac{1}{2}m$

2.

     From energy conservation for system

 Decrease in gravitational potential energy = Increase in spring potential energy + Increase in kinetic energy

3.

$mgx=\frac{1}{2}k{x}^2+\frac{1}{2}m{v}^2$

where $m=5\text{kg},x=\frac{1}{2}\text{meter},k=40\text{N}/\text{m}$

$\Rightarrow 5\times 10\times \frac{1}{2}=\frac{1}{2}\times 40\times {\left(\frac{1}{2}\right)}^2+\frac{1}{2}\times 5v^2$

$\Rightarrow 40=5v^2$

$\Rightarrow v^2=8$

$\Rightarrow v=2\sqrt{2}=2.828 m/s$