Answer to Question #124633 in Physics for Dr. Horus

Question

A 120N force applied horizontally drags a 4kg load along a horizontal table at a uniform velocity of 3m/s.
(i) what is the coefficient of kinetic friction between the load and the table?
(ii) if the applied force is suddenly removed, how much further does the load slide before coming to a stop.

Answer 1

(i) Let's apply the Newton's Second Law of Motion:

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

(ii) We can find the distance that the load slide until it finally stops from the Work - Kinetic Energy theorem:

$$\Delta KE = W,$$ $$KE_f - KE_i = -F_{fr}d,$$ $$\dfrac{1}{2}mv_f^2 - \dfrac{1}{2}mv_i^2 = -\mu_kmgd,$$ $$d = \dfrac{\dfrac{1}{2}mv_f^2 - \dfrac{1}{2}mv_i^2}{-\mu_kmg},$$ $$d = \dfrac{\dfrac{1}{2} \cdot 4 \ kg \cdot [(0 \ \dfrac{m}{s})^2 - (3 \ \dfrac{m}{s})^2 ]}{-3.06 \cdot 4 \ kg \cdot 9.8 \ \dfrac{m}{s^2}} = 0.15 \ m.$$

Answer:

(i) $\mu_k = 3.06.$

(ii) $d = 0.15 \ m.$