The system is shown in Fig. $8.232$ is released from rest with mass $2 k g$ in contact with the ground. Pulley and spring are massless, and friction is absent everywhere. The speed of $5 k g$ block when $2 k g$ block leaves the contact with the ground is (force constant of the spring $k = 40 N m^{- 1}$ and $g = 10 m s^{- 2}$ ).

\sqrt{2} m s^{- 1}

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2 \sqrt{2} m s^{- 1}

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2 m s^{- 1}

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None of these

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Solution

The correct option is D $2 \sqrt{2} m s^{- 1}$
$\begin{matrix} L e t x b e t h e e x t e n s i o n i n s p r i n g w h e n 2 k g m a s s l e a v e s c o n t a c t w i t h t h e g r o u n d k x = 2 k g x = \frac{2 g}{k} = \frac{2 \times 10}{40} = \frac{1}{2} m B y l a w o f c o n s e r v a t i o n o f e n e r g y f o r 5 k g m a s s m g x = \frac{1}{2} k^{2} + \frac{1}{2} m v^{2} v = \sqrt{2 g x - \frac{k x^{2}}{m}} v = \sqrt{2 \times 10 \times \frac{1}{2} - \frac{40}{5} \times \frac{1}{4}} v = 2 \sqrt{2} m / s \end{matrix}$
Hence,
option $(B)$ is correct answer.

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Similar questions

Q. 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}

)

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 $k = 40 N m^{- 1} a n d g = 10 m s^{- 2}$ ).