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

AI Thread Summary
To find the coefficient of friction for a 20.0 kg box being pushed with a force of 105N and accelerating at 3.00 m/s², the net force must first be calculated. The weight of the box is 196N, and the applied force minus the frictional force equals the net force causing the acceleration. The frictional force can be determined by subtracting the net force from the applied force. The coefficient of friction is then calculated using the formula coefficient friction = Ff/Fn, where Ff is the frictional force and Fn is the normal force. Understanding the relationship between applied force, friction, and acceleration is crucial for solving this problem.
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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