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(C) The total mass $M$ of the system placed on the frictionless table is the sum of the infinite series:$M = m + \frac{m}{2!} + \frac{m}{3!} + \ldots

Vedclass · Accepted Answer

$\frac{g}{e}$

Vedclass · Answer

(C) The total mass $M$ of the system placed on the frictionless table is the sum of the infinite series:$M = m + \frac{m}{2!} + \frac{m}{3!} + \ldots + \frac{m}{n!} + \ldots$$M = m \left( 1 + \frac{1}{2!} + \frac{1}{3!} + \ldots + \frac{1}{n!} + \ldots \right)$We know that the expansion of $e$ is $e = 1 + \frac{1}{1!} + \frac{1}{2!} + \frac{1}{3!} + \ldots$Therefore, $1 + \frac{1}{2!} + \frac{1}{3!} + \ldots = e - 1$.Thus, the total mass on the table is $M = m(e - 1)$.Let $a$ be the acceleration of the system. The hanging mass $m$ is pulled by gravity $mg$ and the tension $T$ in the string, while the mass $M$ on the table is pulled by the same tension $T$.For the hanging mass: $mg - T = ma$For the mass on the table: $T = Ma = m(e - 1)a$Substituting $T$ from the second equation into the first:$mg - m(e - 1)a = ma$$g - (e - 1)a = a$$g = a + (e - 1)a = a(1 + e - 1) = ae$$a = \frac{g}{e}$

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