Correct Answer - B
The system ………………..
`F.B.D` of `2 kg` block
When `2kg` block just leaves the contact
`N = 0`
`kx+0=20 implies x = (20)/(40) = (1)/(2)m`
Applying `WET` for the whole system
`w_(g)+w_(s f)= Delta k`
`implies 50xx(1)/(2)-(1)/(2)(40)[((1)/(2))^(2)-0]= (1)/(2)(5)[V^(2)-0]`
`implies V = 2sqrt(2) m//s`