Conceptual question about blocks and friction

[BadSkittles]

Hello everyone,

Two blocks M and m (M is more massive) are sliding freely with the same initial speed
across a floor with friction coefficient μk > 0, and they
come to a stop. Initially there is a distance x between
them. While they are sliding to a stop,

A) The distance between them becomes smaller
B) The distance between them becomes greater
C) The distance between them stays the same

So the equation that I tried with is (Coefficient of Friction) ( Normal Force ) = Friction.

According to this equation, M should have a larger normal force, therefore, a bigger frictional force. But the correct answer is C? Can someone explain this please?!?

 

[QED Andrew]

Your conclusion that the frictional force acting on the block of mass [itex]M[/itex] is larger in magnitude than the frictional force acting on the block of mass [itex]m[/itex] is correct. However, this conclusion does not contradict the fact that the distance between the two blocks stays the same. The block of mass [itex]M[/itex] is acted on by a larger force, but it also has a larger mass. These two factors will cancel out when we use [itex]F = ma[/itex] to calculate the acceleration of the block. The block of mass [itex]M[/itex] undergoes the same acceleration as the block of mass [itex]m[/itex] undergoes.

Consider a more rigorous analysis. The magnitude of the frictional force [itex]F[/itex] acting on the block of mass [itex]m[/itex] is given by

[tex]F = -\mu_k n[/tex]

Therefore, the block of mass [itex]m[/itex] undergoes an acceleration [itex]a[/itex] given by

[tex]a = \dfrac{F}{m} = \dfrac{-\mu_k n}{m} = \dfrac{-\mu_k mg}{m} = -\mu_kg[/tex]

Note the acceleration is independent of the mass of the block. A similar analysis applies to the block of mass [itex]M[/itex].