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

[dorkee]

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.

 

[LowlyPion]

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?

 

[baba89]

To solve for the coefficient of friction, we need to use the equation: coefficient of friction = friction force/normal force. In this case, the normal force is equal to the weight of the box, which is mg, where m is the mass (20.0 kg) and g is the acceleration due to gravity (9.8 m/s^2). So, the normal force is equal to 196N.

Next, we need to find the friction force. We know that the applied force is 105N and the acceleration of the box is 3.00 m/s^2. Using Newton's second law (F=ma), we can calculate the friction force as F = ma = (20.0 kg)(3.00 m/s^2) = 60 N.

Now, we can plug in the values into the equation for coefficient of friction: coefficient of friction = 60 N/196 N = 0.3061.

Therefore, the coefficient of friction for this scenario is 0.3061. This means that the friction force is 30.61% of the normal force, which is relatively high and suggests that there is a lot of resistance between the box and the floor.