1. A skier has an initial velocity of 12.0 m/s, after a while reaches a slope with 18° of inclination and he slides upwards 12.2m before reaching 0 ¿What is the coefficient of friction? "μ" Vi= 12.0 m/s Vf= 0 m/s Distance = 12.2 m 2. Ec = 1/2 m(v²) Wt = Ec(final) - Ec (initial) 3. Wt = (F)(d) = (μ)(N) (D) = (μ)(m)(g)(D) With this in mind: (μ)(m)(g)(D) = Ec(final) this is zero - 1/2 m (v²) The masses go away (μ)(g)(D) = - 1/2 (12²) You clear (μ) (μ)= -72/119.68 That equals (μ) = 0.6 This was way logical when I did it but then I realized I didn't used the degrees of inclination, and I think I have to use them like "(μmgcos18°)(D)" but I'm not that sure. Help? Thanks And I'm from mexico, some terms perhaps are not to clear but I hope to have made myself clear.