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(B) Given: Mass of the rope $M = 5 \,kg$,upward force $F_1 = 100 \,N$,downward force $F_2 = 70 \,N$. Let $g = 10 \,m/s^2$.Using Newton's second law fo

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$85$

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(B) Given: Mass of the rope $M = 5 \,kg$,upward force $F_1 = 100 \,N$,downward force $F_2 = 70 \,N$. Let $g = 10 \,m/s^2$.Using Newton's second law for the whole rope: $F_{net} = Ma$$F_1 - F_2 - Mg = Ma$$100 - 70 - (5 	imes 10) = 5a$$30 - 50 = 5a$$-20 = 5a \implies a = -4 \,m/s^2$ (The negative sign indicates the acceleration is downwards).Now,consider the lower half of the rope (mass $m = M/2 = 2.5 \,kg$). Let $T$ be the tension at the midpoint.The forces acting on the lower half are the tension $T$ upwards,the weight $mg$ downwards,and the force $F_2 = 70 \,N$ downwards.Applying Newton's second law to the lower half: $T - mg - F_2 = ma$$T - (2.5 	imes 10) - 70 = 2.5 	imes (-4)$$T - 25 - 70 = -10$$T - 95 = -10$$T = 95 - 10 = 85 \,N$.

$A$ rope of mass $5 \,kg$ is moving vertically in a vertical position with an upward force of $100 \,N$ acting at the upper end and a downward force of $70 \,N$ acting at the lower end. The tension at the midpoint of the rope is ......... $N$.

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