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Q.

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)

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a

2 ms−1

b

22 ms−1

c

2 ms−1

d

32 ms−1

answer is B.

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Detailed Solution

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=2gK=2×1040=12mNow from conservation of mechanical energy,mgx=12Kx2+12mv22gx−Kx2m=2×10×12−404×5=22 ms−1
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