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(B) The total mass of the system is $M = 7 \ kg + 4 \ kg + 5 \ kg = 16 \ kg$. Applying Newton's second law to the whole system: $F - Mg = Ma$. $200 -

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$112.5$

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(B) The total mass of the system is $M = 7 \ kg + 4 \ kg + 5 \ kg = 16 \ kg$. Applying Newton's second law to the whole system: $F - Mg = Ma$. $200 - 16 	imes 10 = 16 	imes a$. $200 - 160 = 16a$,which gives $40 = 16a$. So,the acceleration is $a = \frac{40}{16} = 2.5 \ m/s^2$. To find the tension $T$ at point $P$ (top of the rope),we consider the system consisting of the $5 \ kg$ block and the $4 \ kg$ rope. The total mass below point $P$ is $m' = 5 \ kg + 4 \ kg = 9 \ kg$. Applying Newton's second law to this subsystem: $T - m'g = m'a$. $T - 9 	imes 10 = 9 	imes 2.5$. $T - 90 = 22.5$. $T = 90 + 22.5 = 112.5 \ N$.

Two blocks of $7 \ kg$ and $5 \ kg$ are connected by a heavy rope of mass $4 \ kg$. An upward force of $200 \ N$ is applied as shown in the diagram. The tension at the top end of the heavy rope at point $P$ is: $(g = 10 \ m/s^2)$ (in $N$)

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