Question

A2.5 kg block is pulled across a horizontal surface that has a coefficient of kinetic friction of 0.60. What is the force of friction exerted by the block?

Answer 1

`\begin{gathered} \text{From the fre}e\text{ body diagram} \\ +\uparrow\Sigma Fy=0 \\ -w=0 \\ N=w \\ \text{but} \\ w=mg \\ m=2.5kg \\ g=9.81m/s^2 \\ N=mg \\ N=(2.5kg)(9.81m/s^2) \\ N=24.5N \\ \text{Also} \\ Ff=N\mu_k \\ \mu_k=0.6 \\ Ff=(24.5N)(0.6) \\ Ff=14.7N \\ \text{The friction force is }14.7N \end{gathered}`