The system shown in figure 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)

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2ms−1

22ms−1

2ms−1

answer is B.

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Detailed Solution

Let x be the extension in the string when the 2 kg block leaves the contact with ground. Then tension in the spring should be equal to the weight of 2 kg block: Kx=2g or x=2gK=2×1040=12mNow from conversion of mechanical energy,mgx=12Kx2+12mv2⇒v=2gx−Kx2m=2×10×12−404×5=22ms−1

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