The pulley and spring are massless,and fricti | vedclass

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(B) For the $2\, kg$ block to leave the ground,the spring force must equal the weight of the $2\, kg$ block: $kx = m_2g \Rightarrow 40x = 2 	imes 10

Vedclass · Accepted Answer

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

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(B) For the $2\, kg$ block to leave the ground,the spring force must equal the weight of the $2\, kg$ block: $kx = m_2g \Rightarrow 40x = 2 	imes 10 \Rightarrow x = 0.5\, m$.Initially,the spring is at its natural length (as the $5\, kg$ block is released from rest). When the $5\, kg$ block descends by $x = 0.5\, m$,the spring stretches by $x = 0.5\, m$.Applying the Conservation of Mechanical Energy $(COME)$ for the system (including the $5\, kg$ block and the spring):Loss in gravitational potential energy of $5\, kg$ block = Gain in kinetic energy of $5\, kg$ block + Gain in elastic potential energy of the spring.$m_1gx = \frac{1}{2}m_1v^2 + \frac{1}{2}kx^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 = 2.5v^2 + 20 	imes 0.25$$25 = 2.5v^2 + 5$$2.5v^2 = 20$$v^2 = 8$$v = \sqrt{8} = 2\sqrt{2}\, m/s$.

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

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