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

Options

A. `sqrt2 ms ^(-1) `
B. `2 sqrt2 ms ^(-1)`
C. `2 m s ^(-1)`
D. `4 sqrt2 ms ^(-1)`

Answer

B

Explanation

Let .x. be the extension in the spring when 2 kg block leaves the contact with ground. Then,
`K x = 2 g`
`or x = (2g)/(K) = (2 xx 10)/(40) = (1)/(2) m`
Now, from conservation of mechnical energy,
`mgx = (1)/(2) K x ^(2) + (1)/(2) mv ^(2)`
or `v = sqrt (2 gx - (Kx ^(2))/( m ))`
`= sqrt (2 xx 10 xx (1)/(2) - (40)/( 4 xx5))= 2 sqrt2 m //s`