Question

If you were exert 20 N of horizontal force while pushing a 10.2 kg box across the floor at a constant velocity what is the coefficient of kinetic friction between the floor and the box

Answer 1

0.1998

Step-by-step explanation

the coefficen of kinetic friction is given by the expression

`\begin{gathered} \mu=(f_f)/(N) \\ wher \\ \mu\text{ is the coefficient of kinetic friction} \\ f_f\text{ is the force of friction} \\ N\text{ is the normal force} \end{gathered}`

Step 1

Free Body Diagram

set the equations of Newton's second law

Newton's law is split into two components along x and y

, that, using our data, give

`\begin{gathered} \Sigma fx=0, \\ \Sigma fx=F-ff=0 \\ so \\ F=ff \\ \text{replacing} \\ 20N=ff \end{gathered}`

Step 2

y-axis

`\begin{gathered} \Sigma fy=0 \\ N-mg=0 \\ so \\ N=mg \\ \text{replacing} \\ N=(10.2\text{ kg)(9.81 }(m)/(s^2)) \\ N=100.062\text{ Newtons} \end{gathered}`

and finally, replace in the equation to find the coefficient of kinetic friction

`\begin{gathered} \mu=(f_f)/(N) \\ \text{replace} \\ \mu=(20N)/(100.062N) \\ \mu=0.1998 \end{gathered}`

therefore, the coefficient of kinetic friction between the floor and the box is

0.1998

I hope this helps you