The boxes of masses 2, {kg} and 8, {kg} are c | vedclass

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(D) Let $m_1 = 8\, {kg}$ and $m_2 = 2\, {kg}$.From the constraint relation,if the $8\, {kg}$ block moves down with acceleration $a$,the $2\, {kg}$ blo

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$0.4$

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(D) Let $m_1 = 8\, {kg}$ and $m_2 = 2\, {kg}$.From the constraint relation,if the $8\, {kg}$ block moves down with acceleration $a$,the $2\, {kg}$ block moves up with acceleration $2a$.The equations of motion are:For $8\, {kg}$ block: $m_1 g - 2T = m_1 a \implies 80 - 2T = 8a \implies 40 - T = 4a \dots (1)$For $2\, {kg}$ block: $T - m_2 g = m_2 (2a) \implies T - 20 = 2(2a) \implies T - 20 = 4a \dots (2)$Adding equations $(1)$ and $(2)$:$(40 - T) + (T - 20) = 4a + 4a$$20 = 8a \implies a = 2.5\, {m/s^2}$.The distance to be covered by the $8\, {kg}$ block is $S = 20\, {cm} = 0.2\, {m}$.Using the kinematic equation $S = ut + \frac{1}{2}at^2$,with $u = 0$:$0.2 = \frac{1}{2} 	imes 2.5 	imes t^2$$0.4 = 2.5 	imes t^2$$t^2 = \frac{0.4}{2.5} = 0.16$$t = \sqrt{0.16} = 0.4\, {s}$.

The boxes of masses $2\, {kg}$ and $8\, {kg}$ are connected by a massless string passing over smooth pulleys as shown in the figure. Calculate the time taken by the box of mass $8\, {kg}$ to strike the ground starting from rest. (Use $g = 10\, {m/s^2}$) (in ${s}$)

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