The system shown in the figure is released fr | vedclass

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(B) Let $x$ be the extension in the spring when the $2\, kg$ block leaves contact with the ground. At this instant,the spring force must balance the w

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$2\sqrt{2}\, m/s$

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(B) Let $x$ be the extension in the spring when the $2\, kg$ block leaves contact with the ground. At this instant,the spring force must balance the weight of the $2\, kg$ block.$Kx = m_2g$$40 	imes x = 2 	imes 10$$x = \frac{20}{40} = 0.5\, m$Now,we apply the principle of conservation of mechanical energy for the $5\, kg$ block as it moves down by a distance $x = 0.5\, m$. The loss in gravitational potential energy of the $5\, kg$ block is equal to the gain in elastic potential energy of the spring plus the gain in kinetic energy of the $5\, kg$ block.$m_1gx = \frac{1}{2}Kx^2 + \frac{1}{2}m_1v^2$$5 	imes 10 	imes 0.5 = \frac{1}{2} 	imes 40 	imes (0.5)^2 + \frac{1}{2} 	imes 5 	imes v^2$$25 = 20 	imes 0.25 + 2.5v^2$$25 = 5 + 2.5v^2$$20 = 2.5v^2$$v^2 = \frac{20}{2.5} = 8$$v = \sqrt{8} = 2\sqrt{2}\, m/s$

The system shown in the figure is released from rest. The pulley and spring are massless,and friction is absent everywhere. The speed of the $5\, kg$ block when the $2\, kg$ block leaves contact with the ground is: (Take the force constant of the spring $K = 40\, N/m$ and $g = 10\, m/s^2$)

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