Question

A box weighing 43 N is pulled horizontally until it slides uniformly over a level floor. If the applied force is 5.5 N, what is the coefficient of friction between the box and the floor?

Answer 1

`\mu=0.13`

Step-by-step explanation:

First, write the summatory of the forces. Note that on the x axis is also 0 because the problem states that the box is moving uniformly (no acceleration):

`\sum F_x= 0 = F- F_f\\\sum F_y= 0 = F_N-W\\`

Now, you know that the force of friction can be writen as the normal force multiplied by the coefficient of friction:

`F_f=\mu N\\`

From the y axis forces and the last expression:

`F_f=\mu W`

Using this expression and the formula for the x axis:

`F- F_f=0\\F-\mu W =0\\\mu = (F)/(W)=(5.5)/(43)=0.13 \\`