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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)

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Solution

The correct option is **B** 2√2 ms−1

Let 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

Let 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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