Two blocks,4, kg and 2, kg,are sliding down a | vedclass

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(B) The system consists of two blocks of masses $m_1 = 4\, kg$ and $m_2 = 2\, kg$ sliding down an inclined plane with angle $	heta = 30^{\circ}$ and

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$2.4$

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(B) The system consists of two blocks of masses $m_1 = 4\, kg$ and $m_2 = 2\, kg$ sliding down an inclined plane with angle $	heta = 30^{\circ}$ and coefficient of kinetic friction $\mu = 0.3$.Since the blocks are in contact and moving together,they will have the same acceleration $a$.The net driving force down the plane is the component of gravity: $F_g = (m_1 + m_2)g \sin 	heta$.The total frictional force opposing the motion is: $f_k = \mu(m_1 + m_2)g \cos 	heta$.Applying Newton's second law for the system: $(m_1 + m_2)a = (m_1 + m_2)g \sin 	heta - \mu(m_1 + m_2)g \cos 	heta$.Dividing by $(m_1 + m_2)$,we get: $a = g(\sin 	heta - \mu \cos 	heta)$.Substituting the values ($g = 9.8\, m/s^2$,$\sin 30^{\circ} = 0.5$,$\cos 30^{\circ} \approx 0.866$):$a = 9.8 	imes (0.5 - 0.3 	imes 0.866) = 9.8 	imes (0.5 - 0.2598) = 9.8 	imes 0.2402 \approx 2.354\, m/s^2$.Rounding to the nearest option,the acceleration is approximately $2.4\, m/s^2$.

Two blocks,$4\, kg$ and $2\, kg$,are sliding down an inclined plane as shown in the figure. The acceleration of the $2\, kg$ block is ........ $m/s^2$.

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