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. (a) `sqrt2ms^-1`
B. (b) `2sqrt2ms^-1`
C. (c) `2ms^-1`
D. (d) `sqrt2ms^-1`

Answer

B

Explanation

Let x be the extension in the spring when `2kg` block leaves the contact with ground. Then tenstion in the spring should be equal to weight of `2kg` block:
`Kx=2g` or `x=(2g)/(K)=(2xx10)/(40)=1/2m`
Now from conservation of mechanical energy,
`mgx=1/2Kx^2+1/2mv^2`
`implies=sqrt(2gx-(Kx^2)/(m))=sqrt(2xx10xx1/2-(40)/(4xx5))=2sqrt2ms^-1`