Answer to Question #253799 in Mechanics | Relativity for taal49d

Question

A box of mass 2.5 kg is pulled up a ramp by a rope at a constant speed. The angle of inclination of the ramp is 15° and the coefficient of kinetic friction of the ramp is 0.28. The rope is parallel to the ramp. Find the tension in the rope. 

Answer 1

Let's apply the Newton's Second Law of Motion in projections on axis $x$- and $y$:

$$F_T-mgsin\theta-F_{fr}=0,$$ $$N-mgcos\theta=0,$$ $$F_T-mgsin\theta-\mu_kN=0,$$ $$F_T-mgsin\theta-\mu_kmgcos\theta=0.$$

From the last equation we can find the tension in the rope:

$$F_T=mg(sin\theta+\mu_kcos\theta),$$ $$F_T=2.5\ kg\times9.8\ \dfrac{m}{s^2}\times(sin15^{\circ}+0.28\times cos15^{\circ})=13\ N.$$