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 k. What minimum constant force has to be applied in the horizontal direction to the bar of mass `m_1` in order to shift the other bar?

Explanation

Let x be the compression in the spring when the bar `m_2` is about to shift. Therefore at this moment spring force on `m_2` is equal to the limiting friction between the bar `m_2` and horizontal floor. Hence
`kx=km_2g` [where `k` is the spring constant (say)] (1)
For the block `m_1` from work-energy theorem: `A=DeltaT=0` for minimum force. (A here includes the work done in streching the spring).
so, `Fx=1/2kx^2-kmgx=0` or `kx/2=F-km_1g` (2)
From (1) and (2),
`F=kg(m_1+m_2/2)`.