Question

A 10 kg box is being pushed to the right with 10.0 N of force while experiencing 8.0 N of friction force. What is the coefficient of friction?

Answer 1

Given,

The mass of the box, m=10 kg

The force with which the box is being pushed, F=10.0 N

The net force acting on the box, Fₙ=8.0 N

The net force acting on the box is given by,

`F_n=F-f`

Where f is the frictional force.

On substituting the known values,

`\begin{gathered} 8.0=10.0-f \\ \Rightarrow f=10.0-8.0=2.0\text{ N} \end{gathered}`

The frictional force is given by the product of the normal force (N) and the coefficient of friction(μ).

And the normal force is given by,

`N=mg`

Thus the frictional force is given by,

`\begin{gathered} f=N\mu \\ =mg\mu \end{gathered}`

Where g is the acceleration due to gravity.

On substituting the known values,

`\begin{gathered} 2.0=10*9.8*\mu \\ \Rightarrow\mu=(2.0)/(10*9.8) \\ =0.02 \end{gathered}`

Thus the coefficient of the friction is 0.02