A78−kg skier is sliding down a ski slope at a constant velocity. the slope makes an angle of 21° above the horizontal direction. (a) neglecting any air resistance, what is the force of kinetic friction acting on the skier? (b) what is the coefficient of kinetic friction between the skis and the snow?

274N 0.41

Explanation:

As he is sliding down in a constant speed then the force that accelerates him (weight) and the force that slows his down (friction) are equal.

then

friction=mass x gravity x sin(21)

Fr=78kg x 9.8m/s2 x sin(21)=274N

friction= coefficient of kinetic friction x normal force of from the slope

Fr= u x 78kg x 9.8m/s2 x cos(21)=274N

Fr= u x 78kg x 9.8m/s2 x cos(21)=274Nu=274/677=0.41

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