Solving for Friction Coefficient: 105N & 3.00 m/s^2

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SUMMARY

The discussion focuses on calculating the coefficient of friction for a 20.0 kg box subjected to a horizontal force of 105N and experiencing an acceleration of 3.00 m/s². The coefficient of friction is determined using the formula coefficient of friction = Ff/Fn. The normal force (Fn) is calculated as 196N, derived from the weight of the box. The difference between the applied force and the frictional force accounts for the box's acceleration, allowing for the calculation of the coefficient of friction.

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dorkee
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Homework Statement


A force of 105N is applied horizontally to a 20.0 kg box to move it across a horizontal floor. If the box has an acceleration of 3.00 m/s^2, find the coefficient friction.

Homework Equations


coefficient friction = Ff/Fn

The Attempt at a Solution


I converted the mass into Newtons so 20.0 kg = 196 N
I have no idea how to do this. I drew a picture, but I don't know what to do with the acceleration.
 
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dorkee said:

Homework Statement


A force of 105N is applied horizontally to a 20.0 kg box to move it across a horizontal floor. If the box has an acceleration of 3.00 m/s^2, find the coefficient friction.


Homework Equations


coefficient friction = Ff/Fn


The Attempt at a Solution


I converted the mass into Newtons so 20.0 kg = 196 N
I have no idea how to do this. I drew a picture, but I don't know what to do with the acceleration.

Ok. What would the acceleration be if there was no friction?

Won't the difference then be how much friction has retarded its motion?
 

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