1. The problem statement, all variables and given/known data A ski starts from rest and slides down a 22o incline 75 m long. a) if the coefficient of friction is 0.090, 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 coefficient of friction, how far will the ski travel along the level? Use energy methods. 2. Relevant equations -Wnc = (KEf - KEo) + (PEf - PEo) 3. The attempt at a solution a)Since the hypotenuse of the incline is 75 m, and the angle relative to the horizontal is 22o, the vertical height is 75 sin22o = 28.1 m. PEf and KEo cancel out, and the equation becomes: PEo = KEf + Wnc That simplifies to: mgh = (1/2)mv2 + umgcos22o -- (u is coefficient of friction) masses cancel out (9.8)(28.1) = (1/2)v2 + (0.09)(9.8)cos22o v = 23.4 m/s -- According to the book, the answer is 21 m/s. What am I doing wrong?? b) KEf = umg * d?