Three particles each of mass m are placed at the three corners of an equilateral triangle of side a. The work done on the system to increase the sides of the triangle to 2a is:

Explanation

The center of mass of an equilateral triangle lies at its geometrical center G. The positions of the mass m,, m, and m, are at positions A, B and C as shown in the Figure. From the given position of the masses, the coordinates of the masses mand m, are easily marked as (0,0) and (1,0) espectively.
To find the position of m, the Pythagoras theorem is applied. A As the ADBC is a right angle triangle,

`BC^(2)=CD^(2)+DB^(2)`
`CD^(2)=BC^(2)-DB^(2)`
`CD^(2)=1^(2)-((1)/(2))^(2)=1-((1)/(4))=(3)/(4)`
`CD=(sqrt(2))/(2)`
The position of mass `m_(3)` is `((1)/(2), (sqrt(3))/(2)) or (0.5, 0.35 sqrt(3))`
X coordinate of center of mass,
`X_(CM)=(m_(1)x_(1)+m_(2)x_(2)+m_(3)x_(3))` `m_(1)+m_(2)m_(3))`
`X_(CM)=((1xx0)+(2xx1)+(3xx0.5))/(1+2+3)=(35)/(6),X_(CM)=(7)/(12)m`
x-coordinate of center of mass,
`Y_(CM)=(m_(1)y_(1)+m_(2)y_(2)+m_(2)m_(3)y_(3))/(m_(1)+m_(2)+m_(3))`
`Y_(CM)=((1xx0)+(2xx0)+(3xx0.5xx sqrt(3)))/(1+2+3)=(1.5 sqrt(3))/(6)`
`Y_(CM)=(sqt(3))/(4)m`
`:.` The coordinates of center of mass Grey) is `x_(CM),y_(CM)` is `((7)/(12), (sqrt(7))/(4))`