Three masses 200 kg, 300 kg, and 400 kg ar | vedclass

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(C) The work done by an external agent is equal to the change in the gravitational potential energy of the system: $W_{ext} = \Delta U = U_f - U_i$.Th

Vedclass · Accepted Answer

$1.74 	imes 10^{-7}$

Vedclass · Answer

(C) The work done by an external agent is equal to the change in the gravitational potential energy of the system: $W_{ext} = \Delta U = U_f - U_i$.The gravitational potential energy of a system of three masses is given by $U = -G \left( \frac{m_1 m_2}{r} + \frac{m_2 m_3}{r} + \frac{m_1 m_3}{r} ight) = -\frac{G}{r} (m_1 m_2 + m_2 m_3 + m_1 m_3)$.Given masses: $m_1 = 200 \ kg$, $m_2 = 300 \ kg$, $m_3 = 400 \ kg$.Sum of products: $(200 	imes 300) + (300 	imes 400) + (200 	imes 400) = 60000 + 120000 + 80000 = 260000 = 2.6 	imes 10^5 \ kg^2$.Initial potential energy $(r_i = 20 \ m)$:$U_i = -\frac{6.67 	imes 10^{-11}}{20} 	imes 2.6 	imes 10^5 = -6.67 	imes 10^{-11} 	imes 0.13 	imes 10^5 = -8.671 	imes 10^{-7} \ J$.Final potential energy $(r_f = 25 \ m)$:$U_f = -\frac{6.67 	imes 10^{-11}}{25} 	imes 2.6 	imes 10^5 = -6.67 	imes 10^{-11} 	imes 0.104 	imes 10^5 = -6.9368 	imes 10^{-7} \ J$.Work done:$W = U_f - U_i = (-6.9368 	imes 10^{-7}) - (-8.671 	imes 10^{-7}) = 1.7342 	imes 10^{-7} \ J \approx 1.74 	imes 10^{-7} \ J$.

Three masses $200 \ kg$, $300 \ kg$, and $400 \ kg$ are placed at the vertices of an equilateral triangle with sides $20 \ m$. They are rearranged on the vertices of a bigger triangle of side $25 \ m$ with the same center. The work done in this process is . . . . . . $J$.

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