What is the coefficient of kinetic friction for a box pulled by a 10N force?

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Homework Help Overview

The problem involves determining the coefficient of kinetic friction for a box being pulled by a force on a horizontal surface, given its mass and acceleration.

Discussion Character

  • Exploratory, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants discuss the application of the formula for the coefficient of friction and how to incorporate acceleration and normal force into their calculations. There are questions about the correctness of the data being used and the interpretation of the equations.

Discussion Status

Some participants have provided guidance on the equations to use, while others are verifying their calculations and discussing potential errors in their math. Multiple interpretations of the problem setup are being explored, but there is no explicit consensus on the final answer.

Contextual Notes

Participants are working with the constraints of the problem statement and are questioning the accuracy of their inputs and calculations, particularly regarding the normal force and the resulting coefficient of friction.

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Homework Statement


A 2.0 kg box is pulled along a horizontal plank by a 10N force parallel to the direction of motion. The rate of acceleration is 0.50 m/s^2. Find the coefficient of kinetic friction.

I think we use the formula: coefficient of friction = F (so 10) / N.
But I am not sure how to use the acceleration and the N variable.


Homework Equations


F - μk mg = ma
10 - 2(9.8)μk = 2 x 0.5

Would this be the correct way to go?

Thanks!
 
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future_vet said:

Homework Statement


A 2.0 kg box is pulled along a horizontal plank by a 10N force parallel to the direction of motion. The rate of acceleration is 0.50 m/s^2. Find the coefficient of kinetic friction.

I think we use the formula: coefficient of friction = F (so 10) / N.
But I am not sure how to use the acceleration and the N variable.


Homework Equations


F - μk mg = ma
10 - 2(9.8)μk = 2 x 0.5

Would this be the correct way to go?

Thanks!
Yes, but it contradicts what you said in the problem statement.
 
Yes, so I use the second equation? Is the data correctly plugged in?
 
future_vet said:
Yes, so I use the second equation? Is the data correctly plugged in?

Yes it is.
 
I am getting -0.005 as the answer... does it sound right? Where should the acceleration be used?

Thanks for all the help! :)
 
future_vet said:
I am getting -0.005 as the answer... does it sound right? Where should the acceleration be used?

Thanks for all the help! :)
you have the correct equation, with the given acceleration entered correctly, but your math is off.
[tex]10 -19.6\mu_k = 2(.5) = 1[/tex]
[tex]-19.6\mu_k = -9[/tex]
[tex]\mu_k = ?[/tex]
 
oops, it's 0.46, right?
 
future_vet said:
oops, it's 0.46, right?
yes, correct.
 

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