A trolley weighing 600 kg moves along an elevated road with an inclination angle of 20 degrees

A trolley weighing 600 kg moves along an elevated road with an inclination angle of 20 degrees under the action of a force of 2.3 kN. What is the coefficient of friction against the surface?

1) Let’s write Newton’s second law for the trolley: mg * sin (20o) – μmg = F (F = 2.3 kN).
2) We express from it the coefficient of friction μ: μ = (mg * sin (20o) – F) / mg = sin (20o) – F / mg.
3) Now we substitute the numerical values and calculate it: μ = sin (20o) – (2.3 * 1000 N) / ((600 kg) * (10 m / s2)) = 0.04.
ANSWER: μ = 0.04.