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°.
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