Asked by Em
A 64.5-kg skier moving horizontally at 4.83 m/s encounters a 16.7° incline.
a) How far up the incline will the skier move before she momentarily stops, ignoring friction?
b) How far up the incline will the skier move if the coefficient of kinetic friction between the skies and snow is 0.103?
a) How far up the incline will the skier move before she momentarily stops, ignoring friction?
b) How far up the incline will the skier move if the coefficient of kinetic friction between the skies and snow is 0.103?
Answers
Answered by
Shuhei
a) v^2/(2g) = h. h/sin(theta) = d
b) v^2/(2*g*mu) = h. h/(mu*sin(theta)) = d
b) v^2/(2*g*mu) = h. h/(mu*sin(theta)) = d
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