Nonconservative Force Question--URGENT! A skier traveling 12 m/s reaches the foot of a steady upward 18 degree incline and glides 12.2 meters up along this slope before coming to rest. What was the average coefficient of friction? Work nonconservative= mechanicalenergyfinal-mechanicalenergyinitial SO Wnc= (1/2mvf^2+mghf)- (1/2mvi^2-mghi) i used Ff= force of friction, and 3.7 coming from sin 18= (x/12.2)- 3.7 is the height of the incline plane. m= mass Ff*12.2= (9.8)(3.7)m-(1/2(12.0^2))m and got Ff(12.2)= 35.74 m so Ff= 2.92m, which means the coefficient of friction, u= 2.92m/9.8mcos18, the masses then cancelling out. my question is, do you have to take the x-component of weight into account at all? does it affect the force of friction and thus the coefficient of friction? or do you just ignore the x-compent of weight, mgsin18? any help would be very much appreciated, i've been debating this all weekend- thank you in advance!