The system shown in Fig. is released from rest with mass 2 kg in contact with the ground. Pulley and spring are massless, and friction is absent everywhere. The speed of 5 kg block when 2 kg block leaves the contact with the ground.is (force constant of the spring `k = 40 Nm^(-1) and g = 10 ms^(-2)`)

Options

A. `sqrt(2) ms^(-1)`
B. `2sqrt(2) ms^(-1)`
C. `2 ms^(-1)`
D. `sqrt(2) ms^(-1)`

Answer

B

Explanation

Let x be the extension in the string when 2 kg block leaves the contact with ground. Then tension in the spring should be equal to weight of 2 kg block:
`Kx = 2g " or" 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`
`= sqrt(2gx - (Kx^2)/(m)) = sqrt(2 xx 10 xx 1/2 - 40/(4 xx 5)) = 2 sqrt(2) ms^(-1)`