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Consider a skier heading down a 9.5° slope. Assume the coefficient of friction for waxed wood on wet snow is μk=0.10 and use a coordinate system in which down the slope is positive.

θ = 9.5 °

Calculate the acceleration of the skier in m/s2.

Find the angle of the slope down which this skier could coast at a constant velocity in degrees.

Jul 16th, 2014

Answers: 0.65 m/s2; 5.7°.

Choose the coordinate system with x-axis directed to the right and the y-axis directed downwards.

The gravity force acting of the skier is W = (0, mg). It can be decomposed into the sum of two forces, FN which is perpendicular to the slope and FT directed along the slope. Their magnitudes are

FN = mg cos θ and FT = mg sin θ, where θ = 9.5°.

The remaining forces acting on the skier are the reaction force -FN and the friction force Fk which is directed along the slope but opposite to the force FT. The magnitude of the friction force is

Fk = μk FN = μk mg cos θ. The resultant of all forces has the magnitude F = FT – Fk =

mg(sin θ – μkcos θ) and creates the acceleration a = F/m = g(sin θ – μkcos θ) =

9.81 (sin 9.5° - 0.10 * cos 9.5°) = 0.65 m/s2.

To answer the second question we need to solve an equation a = 0.

Then sin θ – μkcos θ = 0, tan θ = μk, and θ = tan–1 μk = tan–1 0.10 = 5.7°.

Mar 18th, 2015

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