Question

A 120N force applied horizontally drags a 4kg load along a horizontal table at uniform velocity of 3m/s find the coefficient of kinetic friction between the load and the table

Answer 1

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

$$F_{appl} - F_{fr} = 0,$$ $$F_{appl} = F_{fr} = \mu_kN,$$ $$F_{appl} = \mu_kmg,$$ $$\mu_k = \dfrac{F_{appl}}{mg} = \dfrac{120 \ N}{4 \ kg \cdot 9.8 \ \dfrac{m}{s^2}} = 3.06.$$