System shown in figure is released from rest. Pulley and spring is mass less and friction id absent everywhere. The speed of 5 kg block when 2 kg bock leaves the contact with ground is (Take force constant of spring k = 40 N/M and g = 10 ms^{-2})
A
√2m−1
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B
2√2m−1
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C
2ms=1
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D
4√2m−1
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Solution
The correct option is B2√2m−1 Correct option is (B) When 2 kg block leaves the contact with ground N=0,T=mg=20N
Extension ⇒ Fspring =Tkx=20x=1/2m
Work =Kf−Ki Wravity +ωspring =kf−0Mgx−12Kx2=12mv22√2=V