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Question

Two bars of masses m1 and m2 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. The minimum constant force that has to be applied in the horizontal direction to the bar of mass m1 in order to shift the other bar is Fmin=kg(m1+m2x). Find x.

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Solution

Let x be the compression in the spring when the bar m2 is about to shift. Therefore at this moment spring force on m2 is equal to the limiting friction between the bar m2 and horizontal floor. Hence
kspringx=km2g ......(1) where k is the coefficient of friction.
The spring has to be extended by x to make the block m2 move.
Notice that the bar2 is just about to move, it hasn't started moving. The work done by the constant force
F is a sum of work done on the spring to stretch it by x and moving the bar m1 by x.
Fx=12kspringx2+km1gx
F=12(km2g)+km1g=kg(m1+m22) where we have substituted

kspringx=km2g from (1)
Comparing with Fmin=kg(m1+m2x), we get x=2

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