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Question

# System shown in figure is released from rest. Pulley and spring is massless and friction is absent everywhere. The speed of 5kg block when 2kg block leaves the contact with ground is (Force constant of spring k=40N/m and g=10m/s2 )

A
2m/s
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B
22m/s
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C
2m/s
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D
42m/s
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Solution

## The correct option is B 2√2m/sAs soon as 2 kg block leaves the ground then there will be same tension 'T' in the string such thatT=20N...(i)andT=Fs ...(ii)where,Fs=Kx0...(iii)where,xo→ expansion in spring due to decrease in height of 5 kg block from the ground.from (i), (ii) & (iii)⇒Kxo=20N40xo=20⇒x0=2mApply energy conserrationLoss in (P.E. of block)= gain in (P.E. of spring +K.E of block)⇒mgx0=12Kx20+12mv2⇒5×10×2=12×40×(2)2+12×5×v2100=80+5v22⇒5v22=20⇒v2=405=8⇒v=√2⇒⇒v=2√2ms−1

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