Question

Hulk is helping Natasha pull a 67 - kg box of weapons up a39° inclined plane with a velocity of 6.5 m/s. Hulk is pulling it with a force of 574 N parallel to the 6.19 m incline. The coefficient of kinetic friction is 0.39. What is the speed of the crate at the end of the incline?

Answer 1

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

$$F_{pull}-mgsin\theta-F_{fr}=ma,$$ $$N-mgcos\theta=0.$$

Let's rewrite our equations:

$$F_{pull}-mgsin\theta-\mu_kN=ma,$$ $$N=mgcos\theta.$$

Let's substitute $N$ into the previous equation and find the acceleration of the box:

$$a=\dfrac{F_{pull}-mgsin\theta-\mu_kmgcos\theta}{m}=\dfrac{F_{pull}-mg(sin\theta+\mu_kcos\theta)}{m},$$ $$a=\dfrac{574\ N-67\ kg\cdot9.8\ \dfrac{m}{s^2}\cdot(sin39^{\circ}+0.39\cdot cos39^{\circ})}{67\ kg},$$ $$a=0.804\ \dfrac{m}{s^2}.$$

Finally, we can find the final speed of the box at the end of the incline from the kinematic equation:

$$v_f^2=v_0^2+2as,$$ $$v_f=\sqrt{v_0^2+2as},$$ $$v_f=\sqrt{(6.5\ \dfrac{m}{s})^2+2\cdot0.804\ \dfrac{m}{s^2}\cdot6.19\ m}=7.22\ \dfrac{m}{s}.$$