Stopping a Box of Unknown Mass on Frictionless Floor

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Homework Help Overview

The problem involves a box of unknown mass sliding on a frictionless floor, which encounters a rough region with a specified coefficient of friction. The objective is to determine the shortest length of the rough floor required to stop the box.

Discussion Character

  • Exploratory, Assumption checking

Approaches and Questions Raised

  • The original poster attempts to relate the final velocity, initial velocity, acceleration, and distance using kinematic equations. They question how to determine the normal force and frictional force without knowing the box's weight, seeking alternative methods to approach the problem.

Discussion Status

Some participants suggest that the relationship between force, mass, and acceleration can be utilized, indicating that this might help in resolving the original poster's concerns. However, there is no explicit consensus on the approach taken.

Contextual Notes

The discussion highlights the challenge of working with an unknown mass and the implications of the frictional force in the context of the problem.

TG3
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Homework Statement


A box of unknown mass slides across a frictionless floor with an initial speed of 4.7 m/s. It encounters a rough region where the coefficient of friction is µk = 0.3
Part 1:
What is the shortest length of rough floor which will stop the box?
Part 2:
What is the shortest length of rough floor which will stop the box?

Homework Equations



vf^2 = vi^2 + 2a delta x.
Frictional force = mew times the normal force

The Attempt at a Solution


Final velocity = 0, so
0 = 4.7 ^2 + 2A delta x.
A = the frictional force.
How can I determine the normal force (and after that, the frictional force, and after that, the change in distance) if I don't know the weight of the box? Is there another way to do this problem?
 
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F = ma = μ*m*g

a = μ*g

Does that help?
 
Yes- it certainly does. With that I was able to solve both parts of the problem. I'm writing that formula down for future reference...
 
TG3 said:
Yes- it certainly does. With that I was able to solve both parts of the problem. I'm writing that formula down for future reference...

Better to understand it and remember it forever.
 

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