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,
[tex]F_n=F-f[/tex]Where f is the frictional force.
On substituting the known values,
[tex]\begin{gathered} 8.0=10.0-f \\ \Rightarrow f=10.0-8.0=2.0\text{ N} \end{gathered}[/tex]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,
[tex]N=mg[/tex]Thus the frictional force is given by,
[tex]\begin{gathered} f=N\mu \\ =mg\mu \end{gathered}[/tex]Where g is the acceleration due to gravity.
On substituting the known values,
[tex]\begin{gathered} 2.0=10\times9.8\times\mu \\ \Rightarrow\mu=\frac{2.0}{10\times9.8} \\ =0.02 \end{gathered}[/tex]Thus the coefficient of the friction is 0.02
