Block A weighing 100 , kg rests on a block B | vedclass

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(B) To move block $B$,the applied force $P$ must overcome the frictional forces acting on it.$1$. The friction between block $A$ and block $B$ $(F_{AB

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(B) To move block $B$,the applied force $P$ must overcome the frictional forces acting on it.$1$. The friction between block $A$ and block $B$ $(F_{AB})$ acts on the top surface of block $B$ in the direction opposite to the motion. Since block $A$ is stationary,$F_{AB} = \mu_{AB} \cdot N_A = \mu_{AB} \cdot m_A \cdot g = 0.25 	imes 100 	imes 10 = 250 \, N$.$2$. The friction between block $B$ and the ground $(F_{BS})$ acts on the bottom surface of block $B$. The normal force on the ground is the sum of the weights of both blocks: $N_{ground} = (m_A + m_B)g = (100 + 200) 	imes 10 = 3000 \, N$. Thus,$F_{BS} = \mu_{BS} \cdot N_{ground} = (1/3) 	imes 3000 = 1000 \, N$.$3$. The total force $P$ required is the sum of these two frictional forces: $P = F_{AB} + F_{BS} = 250 + 1000 = 1250 \, N$.

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