A rope of length L and mass M is being pulled | vedclass

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(D) $1$. First,calculate the acceleration of the entire rope. The net force acting on the rope is $F_{net} = F - f_k$,where $f_k = \mu N = \mu Mg = (1

Vedclass · Accepted Answer

$\frac{Mg}{2}$

Vedclass · Answer

(D) $1$. First,calculate the acceleration of the entire rope. The net force acting on the rope is $F_{net} = F - f_k$,where $f_k = \mu N = \mu Mg = (1/2)Mg$. $2$. Thus,$F_{net} = Mg - 0.5Mg = 0.5Mg$. $3$. The acceleration $a$ is given by $a = F_{net} / M = 0.5Mg / M = g/2$. $4$. Now,consider the rear half of the rope (mass $M/2$). The forces acting on this half are the tension $T$ at the midpoint pulling it forward and the kinetic friction $f_k'$ pulling it backward. $5$. The friction on the rear half is $f_k' = \mu (M/2)g = (1/2)(M/2)g = Mg/4$. $6$. Applying Newton's second law to the rear half: $T - f_k' = (M/2)a$. $7$. Substituting the values: $T - Mg/4 = (M/2)(g/2) = Mg/4$. $8$. Therefore,$T = Mg/4 + Mg/4 = Mg/2$.

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