Block A of mass 2 , kg is placed over block B | vedclass

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(D) $1$. Total mass of the system $(A + B) = 2 \, kg + 8 \, kg = 10 \, kg$.$2$. The maximum static friction (limiting friction) between block $B$ and

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Zero

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(D) $1$. Total mass of the system $(A + B) = 2 \, kg + 8 \, kg = 10 \, kg$.$2$. The maximum static friction (limiting friction) between block $B$ and the horizontal floor is given by $f_{max} = \mu_s N$, where $N$ is the normal reaction force from the floor.$3$. The normal reaction $N = (m_A + m_B)g = 10 \, kg 	imes 10 \, ms^{-2} = 100 \, N$.$4$. Therefore, $f_{max} = 0.5 	imes 100 \, N = 50 \, N$.$5$. The applied horizontal force on block $B$ is $F_{app} = 10 \, N$.$6$. Since $F_{app} $7$. Because the system is at rest and no external force is acting on block $A$ to cause relative motion between $A$ and $B$, the static friction force between blocks $A$ and $B$ is zero.

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