Two bars of masses `m_(1)` and `m_(2)` connected by a non`-`deformed light spring rest on a horizontal plane. The coefficient of friction between the bars and the surface is equal to `mu`. What minimum constant force has to be appled in the horizontal direction to the bar of mass `m_(1)` in order to shift the other bar?

Options

A. `(m_(1)+m_(2)/(2))mug`
B. `(m_(2)+m_(1)/(2))mug`
C. `(m_(1)-m_(2)/(2))mug`
D. `(2m_(2)+m_(1)/(2))mug`

Answer

A

Explanation

Let force constant of spring is `k` then `mum_(2)g=kx` & `Fx-mum_(1)g-(1)/(2)kx^(2)=0`
`rArrF=(m_(1)+(m_(2))/(2))mug`