Law of Conservation of Energy (2) A ski starts from rest and slides down a 22 degree incline 75 m long. (a) If the coeffiecent of friction is .090, what is the ski's speed at the base of the incline? (b) If the snow is level at the foot of the incline and has the same coeffiecent of friction, how far will the ski travel along the level? Use energy methods. correct answer: 2400 m ------------------ My thoughts were to use the equation Wnc = KE + PE Seeing as there is no potential energy on the way down I now have Wnc = KE Then I had to find the work acting on the skiis... so... Work(weight) + Work(friction) = (1/2)mv^2 Weights force would be (sin22 x m x g) frictions would be (.090 x N) = (.090 x cos22 mg) I would say the m's cancel out. So then we substitute back into Work(weight) + Work(friction) = (1/2)v^2 ((sin22 x g) x 75m x cos0) + ((.090 x cos22 x g) x 75m x cos180) = 1/2v^2 so I get... 275 + (-61) = .5V^2 214/.5 = v^2 sq root of 428 = v v = 20.6 = 21.0 Answer in Book: 21 m/s Part B Would I use a similar equation but instead of the Wnc = KE it would be Worknc = PE Work nc = mgh ?