How does friction affect the acceleration of a moving box?

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Friction significantly impacts the acceleration of a moving box once it transitions from a frictionless surface to one with kinetic friction. The acceleration can be determined using Newton's Second Law, where the frictional force is expressed as F = -μk * mg, leading to a negative acceleration. The discussion emphasizes the importance of considering the mass distribution of the box over the surfaces when calculating forces. It is noted that the length of the box may influence the calculations, particularly if the distance x exceeds the box's length. Understanding these dynamics is crucial for accurately determining the box's acceleration as it moves across the boundary.
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Homework Statement


The following problem appeared in our test:
A box of mass m is moving with a constant velocity v along a smooth frictionless surface.When the box passes the boundary the frictional force acts on the box.The coefficient of kinetic friction between the box and the surface is μk and the box covers a distance x after crossing the boundary.
Find the dependence of acceleration on x.
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Homework Equations


∑Fx=max

The Attempt at a Solution


I really did not know how to begin the solution.All I did know was that the acceleration would be equal to a=-μkg by Newton's Second law
 
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Why don't you start from the realization you had about the acceleration? How do you relate the acceleration to the traveled distance?
 
By any chance, do they give you the length of the box (like L)?

Chet
 
hacker804 said:
the acceleration would be equal to a=-μkg by Newton's Second law
Not quite. You have canceled out the mass, without considering whether the masses on the two sides of the equation are the same. See Chet's hint/question.
 
Actually I do not remember properly but I think the length of the block was L.
 
hacker804 said:
Actually I do not remember properly but I think the length of the block was L.
I don't think it matters. The lower diagram implies x < L. If you have to allow for x > L as well it gets a little more complicated.
Let's assume x < L for now. Start again with ∑Fx=max. What do you have for the force(s)?
 
The only force is the frictional force F=-uk*mg
 
What fraction of the mass is situated over the frictional surface, and what fraction of the mass is situated over the frictionless surface? What is the normal force over the frictional surface?

Chet
 
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