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 m//s`
B. `2sqrt2 m//s`
C. `2 m//s`
D. `4sqrt2 m//s`

Answer

B

Explanation

Let `x` be the extension in the spring when `2 kg` block leaves the constant the contact with ground.
Then, `kx = 2g`
`x = (2g)/(k) = (2 xx 10)/(40) = (1)/(2) m`
Now from conservation of mechanical energy
`mgx = (1)/(2) kx^(2) + (1)/(2) mv^(2)" " (m = 5 kg)`
`v = sqrt(2 gx - (kx^(2))/(m))`
Substituting the values
`v = sqrt(2 xx 10 xx (1)/(2) - ((40))/(4 xx5)) = 2 sqrt2 m//s`.