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Q.

The system shown in figure 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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a

$2 \sqrt{2} {ms}^{- 1}$

b

$2 {ms}^{- 1}$

c

$\sqrt{2} {ms}^{- 1}$

answer is B.

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Detailed Solution

Let x be the extension in the string when the 2 kg block leaves the contact with ground. Then tension in the spring should be equal to the weight of 2 kg block:

$Kx = 2 g or x = \frac{2 g}{K} = \frac{2 \times 10}{40} = \frac{1}{2} m$

Now from conversion of mechanical energy,

$\begin{array}{l} mgx = \frac{1}{2} {Kx}^{2} + \frac{1}{2} {mv}^{2} \\ \Rightarrow v = \sqrt{2 gx - \frac{{Kx}^{2}}{m}} = \sqrt{2 \times 10 \times \frac{1}{2} - \frac{40}{4 \times 5}} = 2 \sqrt{2} {ms}^{- 1} \end{array}$

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The system shown in figure 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)