The pulley and spring are massless,and fricti | vedclass

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(B) For the $2\,kg$ block to leave the ground,the tension $T$ in the string must equal its weight: $T = m_2 g = 2 	imes 10 = 20\,N$.Since the spring

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

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(B) For the $2\,kg$ block to leave the ground,the tension $T$ in the string must equal its weight: $T = m_2 g = 2 	imes 10 = 20\,N$.Since the spring is connected to the $5\,kg$ block,the spring force $F_s = kx$ must equal the tension $T$ in the string: $kx = 20\,N$.Given $k = 40\,N/m$,we have $40x = 20$,which gives the extension $x = 0.5\,m$.We apply the Conservation of Mechanical Energy $(COME)$ for the system consisting of the $5\,kg$ block and the spring. The initial state is at rest,and the final state is when the $5\,kg$ block has moved down by $x = 0.5\,m$.The loss in gravitational potential energy of the $5\,kg$ block equals the gain in kinetic energy of the $5\,kg$ block plus the gain in elastic potential energy of the spring:$m_1 g x = \frac{1}{2} m_1 v^2 + \frac{1}{2} k x^2$$5 	imes 10 	imes 0.5 = \frac{1}{2} 	imes 5 	imes v^2 + \frac{1}{2} 	imes 40 	imes (0.5)^2$$25 = 2.5 v^2 + 20 	imes 0.25$$25 = 2.5 v^2 + 5$$2.5 v^2 = 20$$v^2 = 8$$v = \sqrt{8} = 2\sqrt{2}\,m/s$.

The pulley and spring are massless,and friction is absent everywhere. The $5\,kg$ block is released from rest. 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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