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Q.

In the system shown the 5 kg block is released from rest. The pulley and spring are massless and friction is absent everywhere. The force constant of spring is 40 Nm. The speed of 5 kg block when 2 kg block leaves contact with ground is (Take g = 10 ms-2)

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a

2ms−1

b

22ms−1

c

2ms−1

d

42ms−1

answer is B.

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Detailed Solution

Let x be the extension in the spring when 2 kg block leaves the contact with ground. Then,     kx=2g  ⇒ x    =2gk=2×1040=12mNow by Law of Conservation of energy, we havemgx=12kx2+12mv2         { where m=5kg}⇒v=2gx−kx2m⇒v=2×10×12−404×5=22ms−1
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