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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 )
1206667_bc5075b6e45a40aeb3a0beb610b81e66.png

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 22m/s
As soon as 2 kg block leaves the ground then there will be same tension 'T' in the string such that
T=20N...(i)
and
T=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=20N
40xo=20
x0=2m
Apply energy conserration
Loss 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×v2
100=80+5v22
5v22=20
v2=405=8
v=2
v=22ms1

1176138_1206667_ans_7f2d98738797486eb66d88c49d7fa084.jpg

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