Two blocks of masses m and 2m kept on a frict | vedclass

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(C) Let the acceleration of the system be $a$. The net force acting on the system is $F_{net} = F_2 - F_1 = 7.5t - 4.2t = 3.3t 	ext{ N}$.The total ma

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$2$

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(C) Let the acceleration of the system be $a$. The net force acting on the system is $F_{net} = F_2 - F_1 = 7.5t - 4.2t = 3.3t 	ext{ N}$.The total mass of the system is $M = m + 2m = 3m$.Using Newton's second law,$F_{net} = Ma$,we get $3.3t = (3m)a$,so $a = \frac{1.1t}{m}$.Now,consider the block of mass $2m$. The forces acting on it are $F_2$ in the forward direction and tension $T$ in the backward direction.Applying Newton's second law to this block: $F_2 - T = (2m)a$.Substituting the values: $7.5t - T = 2m \left( \frac{1.1t}{m} ight) = 2.2t$.Thus,$T = 7.5t - 2.2t = 5.3t$.Given that the tension $T = 10.6 	ext{ N}$,we have $5.3t = 10.6$.Solving for $t$,we get $t = \frac{10.6}{5.3} = 2 	ext{ s}$.

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