A 65 kg water skier is pulled up a 15° inclined by a role parallel to the incline with a tension of 500N. The coefficient of kinetic friction is .25. What are the magnitude and direction of the skier's acceleration?

Question

A 65 kg water skier is pulled up a 15° inclined by a role  parallel to the incline with a tension of 500N. The coefficient of kinetic friction is .25. What are the magnitude and direction of the skier's acceleration?

I found the weight its 637.65
I found it by multiplying the mass by gravity: 65*9.81=637.65

I don't know how to find the magnitude and direction of the acceleration

Please help.

Henry · Answer

m*g = 65 * 9.8 = 637 N. = Force of skier.

Fp = 637*sin15 = 164.9 N. = Force parallel to the incline.

Fn = 637*cos15 = 615.3 N. = Normal force
= Force perpendicular to the incline.

Fk = u*Fn = 0.25 * 615.3 = 153.8 N. =
Force of kinetic friction.

a = (Fap-Fp-Fk)/m.
Fap = 500 N. = Force applied.
Solve for a.
Direction: Up the ramp.