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(B) The system of blocks $A$ and $B$ is moving with a constant velocity of $10 \, m/s$. Since the velocity is constant,the acceleration of the system

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$20$

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(B) The system of blocks $A$ and $B$ is moving with a constant velocity of $10 \, m/s$. Since the velocity is constant,the acceleration of the system is $a = 0 \, m/s^2$. According to Newton's second law of motion,the net external force acting on the system in the horizontal direction must be zero. Let $F_{ext} = 20 \, N$ be the applied force on block $B$ in the rightward direction. Let $f_g$ be the friction force applied by the ground on block $B$ acting in the leftward direction. For the system to be in equilibrium (constant velocity),the net force in the horizontal direction must be zero: $F_{ext} - f_g = 0$ $20 \, N - f_g = 0$ $f_g = 20 \, N$. Therefore,the friction force applied by the ground on block $B$ is $20 \, N$.

The blocks shown in the figure move with a constant velocity of $10 \, m/s$ towards the right. All surfaces in contact are rough. The friction force applied by the ground on block $B$ is ........ $N$.

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