Question

A 425N force is applied horizontally to a 33 kg box on a flat table, if the acceleration of the box is 5 m/s2 what is the coefficient of friction between the box and the table?

Answer 1

Let's apply the Newton's Second Law of Motion:

$$F_{appl}-F_{fr}=ma,$$ $$F_{appl}-\mu mg=ma,$$ $$\mu=\dfrac{F_{appl}-ma}{mg},$$ $$\mu=\dfrac{425\ N-33\ kg\cdot 5\ \dfrac{m}{s^2}}{33\ kg\cdot 9.8\ \dfrac{m}{s^2}}=0.80$$

Answer:

$\mu=0.80$