System shown in figure is released from rest. Pulley and spring are massless and friction is absent everywhere. The speed of 5kg block when 2kg block leaves contact with ground is $(K = 40 N m^{- 1}, g = 10 m s^{- 2})$ ______ m/s.

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answer is 2.828.

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Detailed Solution

For 2kg: $2 g = k x \Rightarrow x = \frac{2 \times 10}{40} = \frac{1}{2} m$
Loss of P.E of 5kg block = gain of elastic potential energy of spring+ gain of kinetic energy of 5kg
$\begin{array}{l} \Rightarrow 5 \times g x = \frac{1}{2} k x^{2} + \frac{1}{2} m v^{2} \\ \Rightarrow 5 \times 10 \times \frac{1}{2} = \frac{1}{2} \times 40 \times \frac{1}{4} + \frac{1}{2} \times 5 \times v^{2} \\ \Rightarrow 4 = \frac{v^{2}}{2} \\ \Rightarrow v = 2 \sqrt{2} = 2 \times 1.414 = 2.828 {ms}^{- 1} \end{array}$