A uniform rope of mass 1.0, kg is connected t | vedclass

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(C) The total mass of the system is $M = m_{	ext{rope}} + m_{	ext{box}} = 1.0\, kg + 2.0\, kg = 3.0\, kg$.The acceleration $a$ of the system is give

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$5$

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(C) The total mass of the system is $M = m_{	ext{rope}} + m_{	ext{box}} = 1.0\, kg + 2.0\, kg = 3.0\, kg$.The acceleration $a$ of the system is given by $a = \frac{F}{M} = \frac{6\, N}{3.0\, kg} = 2\, m/s^2$.To find the tension at the midpoint of the rope,we consider the system consisting of the box and the half of the rope attached to it. The mass of this part is $m' = m_{	ext{box}} + \frac{1}{2} m_{	ext{rope}} = 2.0\, kg + 0.5\, kg = 2.5\, kg$.The tension $T$ at the midpoint is the force required to accelerate this combined mass $m'$ at $a = 2\, m/s^2$.$T = m' 	imes a = 2.5\, kg 	imes 2\, m/s^2 = 5\, N$.

$A$ uniform rope of mass $1.0\, kg$ is connected to a box of mass $2.0\, kg$,which is placed on a smooth horizontal surface. The free end of the rope is pulled horizontally by a force of $6\, N$. Find the tension at the midpoint of the rope in $N$.

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