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(B) First,consider the entire system as a single body to find the acceleration $a$.The total mass of the system is $M = 7\,kg + 4\,kg + 5\,kg = 16\,kg

Vedclass · Accepted Answer

$112.5$

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(B) First,consider the entire system as a single body to find the acceleration $a$.The total mass of the system is $M = 7\,kg + 4\,kg + 5\,kg = 16\,kg$.The net upward force is $F = 200\,N$.The total downward force due to gravity is $Mg = 16 	imes 10 = 160\,N$.Using Newton's second law,$F_{net} = Ma$:$200 - 160 = 16a$$40 = 16a$$a = \frac{40}{16} = 2.5\,m/s^2$.Now,to find the tension $T$ at point $P$ (the top of the rope),consider the system consisting of the $5\,kg$ block and the $4\,kg$ rope together.The total mass below point $P$ is $m' = 5\,kg + 4\,kg = 9\,kg$.The forces acting on this subsystem are the tension $T$ upwards and the weight $m'g$ downwards.Using Newton's second law for 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 of the heavy rope at point $P$ is ....... $N$ $(g = 10\,m/s^2)$.

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