A block of mass 2M is attached to a massless | vedclass

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Question

(C) Let $T$ be the tension in the string connecting the two hanging blocks. For the movable pulley,the tension in the string connected to the block of

Vedclass · Accepted Answer

$a_2 - a_1 = a_1 - a_3$

Vedclass · Answer

(C) Let $T$ be the tension in the string connecting the two hanging blocks. For the movable pulley,the tension in the string connected to the block of mass $2M$ is $2T$. The equation of motion for the block of mass $2M$ is $2T - kx = 2Ma_1$. For the hanging system,the relative acceleration of the blocks $M$ and $2M$ with respect to the movable pulley is $a_{rel}$. The equations are $Mg - T = M(a_1 - a_{rel})$ and $2Mg - T = 2M(a_1 + a_{rel})$. Solving these gives $T = \frac{4}{3}Mg$ and $a_{rel} = \frac{g}{3}$. The acceleration of block $M$ is $a_2 = a_1 + a_{rel}$ and for $2M$ is $a_3 = a_1 - a_{rel}$. Thus,$a_2 - a_1 = a_1 - a_3 = a_{rel} = \frac{g}{3}$. Substituting $T$ into the first equation: $2(\frac{4}{3}Mg) - kx = 2Ma_1 \implies a_1 = \frac{4g}{3} - \frac{kx}{2M}$. At maximum extension $x_0$,$a_1 = 0$,so $x_0 = \frac{8Mg}{3k}$. Option $A$ is incorrect. For option $C$,$a_2 - a_1 = a_{rel}$ and $a_1 - a_3 = a_{rel}$,so $a_2 - a_1 = a_1 - a_3$ is correct. At $x = \frac{x_0}{4}$,$a_1 = \frac{4g}{3} - \frac{k(8Mg/12k)}{2M} = \frac{4g}{3} - \frac{g}{3} = g$. Option $D$ is incorrect.

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