Three masses m, 2m and 3m are arranged in two | vedclass

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(D) The work done by an external agent is equal to the change in the gravitational potential energy of the system: $W = U_f - U_i$.For configuration $

Vedclass · Accepted Answer

$-\frac{G m^2}{a}\left[6-\frac{6}{\sqrt{2}}\right]$

Vedclass · Answer

(D) The work done by an external agent is equal to the change in the gravitational potential energy of the system: $W = U_f - U_i$.For configuration $1$ (right-angled triangle with sides $a, a, \sqrt{2}a$):$U_i = -\frac{G(m)(2m)}{a} - \frac{G(m)(3m)}{a} - \frac{G(2m)(3m)}{\sqrt{2}a} = -\frac{G m^2}{a} \left( 2 + 3 + \frac{6}{\sqrt{2}} \right) = -\frac{G m^2}{a} \left( 5 + \frac{6}{\sqrt{2}} \right)$.For configuration $2$ (equilateral triangle with all sides $a$):$U_f = -\frac{G(m)(3m)}{a} - \frac{G(m)(2m)}{a} - \frac{G(2m)(3m)}{a} = -\frac{G m^2}{a} (3 + 2 + 6) = -\frac{11 G m^2}{a}$.Work done $W = U_f - U_i = -\frac{11 G m^2}{a} - \left( -\frac{G m^2}{a} \left( 5 + \frac{6}{\sqrt{2}} \right) \right) = \frac{G m^2}{a} \left( -11 + 5 + \frac{6}{\sqrt{2}} \right) = \frac{G m^2}{a} \left( \frac{6}{\sqrt{2}} - 6 \right) = -\frac{G m^2}{a} \left( 6 - \frac{6}{\sqrt{2}} \right)$.

Three masses $m, 2m$ and $3m$ are arranged in two triangular configurations as shown in figure $1$ and figure $2$. The work done by an external agent in changing the configuration from figure $1$ to figure $2$ is:

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