Posted by Anonymous on .
An 60 kg skier is sliding down a ski slope at a constant velocity. The slope makes an angle of 13° above the horizontal direction.
(a) Ignoring any air resistance, what is the force of kinetic friction acting on the skier? in N
(b) What is the coefficient of kinetic friction between the skis and the snow?

PHYSICSSSSSS please help 
Henry,
Ws = mg = 60kg * 9.8N/kg = 588 N. = Wt.
of skier.
Fs = 588 N. @ 13 Deg. = Force of skier.
Fp = 588*sin13 = 132.3 N. = Force parallel to incline.
Fv = 588*cos13 = 572.9 N. = Force perpendicular to incline.
a. Fp Fk = 0, Because a = 0.
132.3  Fk = 0,
Fk = 132.3 N. = Force of kinetic friction.
b. u*Fv = Fk = 132.3 N,
u*572.9 = 132,3,
u = 132.3 / 572.9=0.232. = coefficient
of frictin. 
PHYSICSSSSSS please help 
Anonymous,
THNKS :)