224k views
2 votes
A 5kg block slides across a horizontal table surface with a coefficient of friction of 0.21. What is the friction force the block experiences?

1 Answer

2 votes

\begin{gathered} m=5\operatorname{kg},g=9.81m/s^2 \\ \mu_k=0.21 \\ Ff=\text{?} \\ Ff=\text{ N}\cdot\mu_k \\ \text{From the fr}ee\text{ body diagram} \\ \uparrow+\Sigma Fy=0 \\ N-mg=0 \\ N=mg \\ \text{Hence} \\ Ff=\text{ N}\cdot\mu_k \\ Ff=\text{ }mg\cdot\mu_k \\ Ff=(5\operatorname{kg})(9.81m/s^2)(0.21) \\ Ff=10.3N \\ \text{The friction force is 10.3N} \end{gathered}

A 5kg block slides across a horizontal table surface with a coefficient of friction-example-1
A 5kg block slides across a horizontal table surface with a coefficient of friction-example-2
User Kaan Burak Sener
by
5.3k points