The system shown in Fig. is released from rest with mass 2 kg in contact with the ground. Pulley and spring are massless, and friction is absent everywhere. The speed of 5 kg block when 2 kg block leaves the contact with the ground is (force constant of the spring k = 40 N $m^{- 1}$ and g = 10 m $s^{- 2}$ )

Question

Infinity Learn · Accepted Answer

Let x be the extension in the string when $$2$$ kg block leaves the contact with ground. Then tension in the spring should be equal to weight of $$2$$ kg block:Kx $$=$$ $$2g$$ or $$x=2gK=2$$×$$1040=12mNow$$ from conservation of mechanical energy,$$mgx=12Kx2+12mv22gx$$−$$Kx2m=2$$×$$10$$×$$12$$−$$404$$×$$5=22$$ ms−$$1$$

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

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