Asked by Anonymous
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?
(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?
Answers
Answered by
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.
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.
Answered by
Anonymous
THNKS :)
There are no AI answers yet. The ability to request AI answers is coming soon!
Submit Your Answer
We prioritize human answers over AI answers.
If you are human, and you can answer this question, please submit your answer.