a 68kg skier is speeding down a ski slope that makes a 32 angle with the horizontal. The co-efficient of friction between the skies and the snow is 0.05. What is the acceleration of the skier?

Question

a 68kg skier is speeding down a ski slope that makes a 32 angle with the horizontal. The co-efficient of friction between the skies and the snow is 0.05. What is the acceleration of the skier?
I want to understand how to do this

Henry · Answer

m*g = 68kg * 9.8N./kg = 666.4 N. = Wt. of skier.

Fp = 666.4*sin32 = 353.1 N. = Force parallel with incline.

Fn = 666.4*cos32 = 565.1 N. = Normal =
Force perpendicular to the incline.

Fk = u*Fn = 0.05 * 565.1 = 28.26 N. =
Force of kinetic friction.

Fp-Fk = m*a
a=(Fp-Fk)/m =(353.1-28.26)/68=4.78m/s^2

Note: Fp-Fk = Net force.