How does friction affect the acceleration of a moving box?

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Homework Help Overview

The problem involves a box of mass m moving with a constant velocity along a frictionless surface, which then encounters friction upon crossing a boundary. The task is to determine how acceleration depends on the distance x covered after the boundary, considering the coefficient of kinetic friction μk.

Discussion Character

  • Exploratory, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • Participants discuss the relationship between acceleration and distance traveled, with some questioning the initial assumptions about mass and forces involved. There are attempts to clarify the role of the box's length in the problem.

Discussion Status

Participants are exploring different interpretations of the problem, particularly regarding the effects of friction and the setup of forces. Some guidance has been offered to reconsider the relationship between mass and acceleration, but no consensus has been reached.

Contextual Notes

There is uncertainty regarding the length of the box and its relevance to the problem, with some participants suggesting it may complicate the analysis if x exceeds L.

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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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