A block A with mass 100, kg is resting on ano | vedclass

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(C) To start moving block $B$,the applied force $F$ must overcome the frictional forces acting on it.There are two frictional forces opposing the moti

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$1100$

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(C) To start moving block $B$,the applied force $F$ must overcome the frictional forces acting on it.There are two frictional forces opposing the motion of block $B$:$1$. The kinetic friction between block $A$ and block $B$ $(f_{AB})$,which acts on $B$ in the direction opposite to its motion.$f_{AB} = \mu_{AB} \cdot N_A = \mu_{AB} \cdot m_A \cdot g = 0.2 	imes 100 	imes 10 = 200\, N$.$2$. The kinetic friction between block $B$ and the ground $(f_{BG})$,which acts on $B$ in the direction opposite to its motion.The normal force on the ground is the sum of the weights of both blocks: $N_G = (m_A + m_B)g = (100 + 200) 	imes 10 = 3000\, N$.$f_{BG} = \mu_{BG} \cdot N_G = 0.3 	imes 3000 = 900\, N$.Therefore,the minimum force $F$ required to move block $B$ is the sum of these two frictional forces:$F = f_{AB} + f_{BG} = 200\, N + 900\, N = 1100\, N$.

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