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 5 kg block when 2 kg block leaves the contact with ground is (Take force constant of spring K=40 N/m and g=10 m/s2,2=1.414)

1.  2 kgblock will leave the contact if T=mg(m=2 kg)Also, T=kx0⇒x0=12m2.     From energy conservation for system Decrease in gravitational potential energy = Increase in spring potential energy + Increase in kinetic energy3.mgx=12kx2+12mv2where m=5 kg,x=12meter,k=40N/m⇒5×10×12=12×40×(12)2+12×5v2⇒40=5v2⇒v2=8⇒v=22Therefore, the correct answer is 2.83