Conceptual question about blocks and friction

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SUMMARY

In the discussion, two blocks, M and m, slide on a frictional surface with coefficient μk, initially separated by distance x. Despite block M having a larger mass and thus a greater frictional force acting on it, both blocks experience the same acceleration due to friction, leading to the conclusion that the distance between them remains constant as they come to a stop. The frictional force is calculated using the equation F = -μk n, demonstrating that acceleration is independent of mass.

PREREQUISITES
  • Understanding of Newton's Second Law (F = ma)
  • Knowledge of frictional forces and coefficients (μk)
  • Basic concepts of mass and acceleration
  • Familiarity with free body diagrams
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  • Study the effects of varying coefficients of friction on motion
  • Learn about free body diagram analysis for multiple objects
  • Explore the relationship between mass, force, and acceleration in different contexts
  • Investigate real-world applications of friction in engineering and physics
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BadSkittles
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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?!?
 
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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].
 
Last edited:

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