Question

A 60.0-kg skier with an initial speed of 12.0 m/s coasts up a 2.50-m-high rise

a. Find her final speed at the top, given that the coefficient of friction between her skis and the snow is 0.0800. (Hint: Find the distance traveled up the incline assuming a straight-line path as shown in the figure.)

b. How much work does friction do on the skier?

Answer 1

(a) Let's apply the law of conservation of energy:

$$KE_i+PE_i+W_{NC}=KE_f+PE_f,$$ $$\dfrac{1}{2}mv_i^2+0+(-F_{fr}d)=\dfrac{1}{2}mv_f^2+mgh.$$

We can find the force of friction as follows:

$$F_{fr}=\mu_kN=\mu_kmgcos\theta.$$

We can find the distance traveled up the incline from the geometry:

$$sin\theta=\dfrac{h}{d},$$ $$d=\dfrac{h}{sin\theta}.$$

Then, we get:

$$\dfrac{1}{2}mv_i^2-\mu_kmgcos\theta\cdot\dfrac{h}{sin\theta}=\dfrac{1}{2}mv_f^2+mgh,$$ $$v_f=\sqrt{v_i^2-2gh(\mu_k cot\theta+1)},$$ $$v_f=\sqrt{(12\ \dfrac{m}{s^2})^2-2\cdot9.8\ \dfrac{m}{s^2}\cdot2.5\ m\cdot(0.08\cdot cot35^{\circ}+1)},$$ $$v_f=9.45\ \dfrac{m}{s}.$$

(b) We can find the work done by the friction force on skier as follows:

$$W_{fr}=-F_{fr}d=-\mu_kmghcot\theta,$$ $$W_{fr}=-0.08\cdot60\ kg\cdot9.8\ \dfrac{m}{s^2}\cdot2.5\ m\cdot cot35^{\circ}=-168\ N.$$