A rope of mass 5,kg is moving vertically in a | vedclass

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(A) Let the mass of the rope be $M = 5\,kg$. The net force acting on the rope is $F_{net} = F_{up} - F_{down} - Mg = 100 - 70 - (5 	imes 10) = 100 -

Vedclass · Accepted Answer

$85$

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(A) Let the mass of the rope be $M = 5\,kg$. The net force acting on the rope is $F_{net} = F_{up} - F_{down} - Mg = 100 - 70 - (5 	imes 10) = 100 - 70 - 50 = -20\,N$.Since the net force is $-20\,N$,the acceleration $a = F_{net} / M = -20 / 5 = -4\,m/s^2$ (downwards).Consider the upper half of the rope (mass $m = 2.5\,kg$). Let $T$ be the tension at the midpoint.The forces acting on the upper half are the upward force $100\,N$,the weight $mg = 2.5 	imes 10 = 25\,N$ downwards,and the tension $T$ downwards.Applying Newton's second law: $100 - T - 25 = m 	imes a$.$75 - T = 2.5 	imes (-4)$.$75 - T = -10$.$T = 75 + 10 = 85\,N$.

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

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