The velocity at which the 6 text{ kg} mass (s | vedclass

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(A) Using the principle of conservation of mechanical energy,the loss in potential energy of the $6 	ext{ kg}$ mass is equal to the gain in kinetic e

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(A) Using the principle of conservation of mechanical energy,the loss in potential energy of the $6 	ext{ kg}$ mass is equal to the gain in kinetic energy of the system (both $6 	ext{ kg}$ and $2 	ext{ kg}$ masses) plus the gain in potential energy of the $2 	ext{ kg}$ mass.Let $m_1 = 6 	ext{ kg}$ and $m_2 = 2 	ext{ kg}$. When the $6 	ext{ kg}$ mass falls by $h = 6 	ext{ m}$,the $2 	ext{ kg}$ mass rises by $h = 6 	ext{ m}$.Loss in potential energy of $m_1 = m_1 g h = 6 	imes 10 	imes 6 = 360 	ext{ J}$.Gain in potential energy of $m_2 = m_2 g h = 2 	imes 10 	imes 6 = 120 	ext{ J}$.Gain in kinetic energy of the system = $\frac{1}{2}(m_1 + m_2)v^2 = \frac{1}{2}(6 + 2)v^2 = 4v^2$.By conservation of energy: $m_1 g h = m_2 g h + \frac{1}{2}(m_1 + m_2)v^2$.$360 = 120 + 4v^2$.$240 = 4v^2$.$v^2 = 60$.$v = \sqrt{60} \approx 7.746 	ext{ m/s}$.

The velocity at which the $6 \text{ kg}$ mass (shown in the figure) strikes the ground when it is released from a height of $6 \text{ m}$ above the ground is . . . . . . $\text{m/s}$. Assume the pulley is massless and the string is light and inextensible. (Take $g = 10 \text{ m/s}^2$)

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