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