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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 tension in the string equals t

Vedclass · Accepted Answer

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

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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 tension in the string equals the weight of the $2 \, kg$ block.$kx = m_2 g$$x = \frac{m_2 g}{k} = \frac{2 	imes 10}{40} = 0.5 \, m$Now,apply the principle of conservation of mechanical energy for the $5 \, kg$ block as it moves down by a distance $x = 0.5 \, m$:Loss in gravitational potential energy = Gain in spring potential energy + Gain in kinetic energy$m_1 g x = \frac{1}{2} k x^2 + \frac{1}{2} m_1 v^2$$5 	imes 10 	imes 0.5 = \frac{1}{2} 	imes 40 	imes (0.5)^2 + \frac{1}{2} 	imes 5 	imes v^2$$25 = 20 	imes 0.25 + 2.5 v^2$$25 = 5 + 2.5 v^2$$20 = 2.5 v^2$$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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