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Question

# The system shown in the figure is released from rest with the mass 2 kg in contact with the ground. The pulley and spring are massless, and friction is absent everywhere. Then, find the speed of the 5 kg block when the 2 kg block leaves contact with the ground (force constant of the spring k=40 Nm−1 and g=10 ms−2)

A
2 ms1
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B
22 ms1
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C
2 ms1
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D
23 ms1
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Solution

## The correct option is B 2√2 ms−1Let x be the extension in the string when 2 kg block leaves contact with the ground. For 2 kg block just breaking contact with the ground, Normal reaction N=0 i.e T=mg=20 N Since the string is joined with one end of the spring and both spring and string are massless, so tension will be the same everywhere i.e T=spring force T=kx ...(i) ⇒kx=T=20 N ∴x=20k=2040=0.5 m Applying mechanical energy conservation on the system (blocks+spring+string): (since the internal forces are non-dissipative and work done by external force from ceiling zero, and external gravitational force is also conservative) ∴Loss in gravitational PE of 5 kg block=Gain in elastic PE of spring+Gain in KE of 5 kg block i.e Mgx=12kx2+12Mv2 Here M=5 kg ⇒5×10×0.5=12×40×0.52+12×5×v2 ⇒50=10+5v2 ∴v=√8=2√2 m/s

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