Given data:
* The mass of Mr. Mangan is 86 kg.
* The friction force acting on the Mr. Mangan is 150 N.
Solution:
The normal force acting on the Mr. Mangan is,
[tex]F_N=mg[/tex]where m is the mass of Mr. Mangan, and g is the acceleration due to gravity,
Substituting the known values,
[tex]\begin{gathered} F_N=86\times9.8 \\ F_N=842.8\text{ N} \end{gathered}[/tex]The frictional force acting on the Mr. Mangan in terms of coefficient of friciton is,
[tex]F_{\text{k}}=\mu_kF_N[/tex][tex]\text{where }\mu_k\text{ is the coefficient of friction,}[/tex]Substituting the known values,
[tex]\begin{gathered} 150=\mu_k\times842.8 \\ \mu_k=\frac{150}{842.8} \\ \mu_k=0.18 \end{gathered}[/tex]Thus, the coefficient of kinetic friction is 0.18.
