System shown in figure is released from rest . Pulley and spring is mass less and friction is absent everywhere. The speed of `5 kg` block when `2 kg` block leaves the constant of with ground is (force constant of spring `k = 40 N//m and g = 10 m//s^(2))`

Options

A. `sqrt(2) m//s`
B. `2sqrt(2) m//s`
C. `2 m//s`
D. None of these

Answer

2

Explanation

Let x be the extension is spring when 2 kg block leaves the constant with ground. Then kx = 2g.
`x = (2g)/(k)= (2xx 10)/(40) = (1)/(2)m`
Now from conservation of energy ,
`mgx = (1)/(2) Kx^(2) +(1)/(2)mv^(2)`
`rArr v=sqrt(2gx-(kx^(2))/(m))`
`=2sqrt(2)m//s`