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(D) Let the acceleration of the system be $a$. The total mass of the system is $M = 5 	ext{ kg} + 3 	ext{ kg} + 1 	ext{ kg} = 9 	ext{ kg}$.The net

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$\frac{10g}{9}$

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(D) Let the acceleration of the system be $a$. The total mass of the system is $M = 5 	ext{ kg} + 3 	ext{ kg} + 1 	ext{ kg} = 9 	ext{ kg}$.The net driving force is the difference in weights: $F_{	ext{net}} = (5g) - (3g + 1g) = 5g - 4g = g$.Using Newton's second law,$F_{	ext{net}} = Ma$,so $g = 9a$,which gives $a = \frac{g}{9}$.Now,consider the free body diagram of block $B$ ($1 	ext{ kg}$ mass). The forces acting on it are its weight $(1g)$ downwards and the tension $T_2$ upwards. Since the system accelerates such that block $B$ moves upwards,we have:$T_2 - 1g = 1a$$T_2 = g + a = g + \frac{g}{9} = \frac{10g}{9}$.Thus,the tension in the string connecting $A$ and $B$ is $\frac{10g}{9}$.

In the adjoining figure,the tension in the string connecting blocks $A$ and $B$ is:

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