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.
Answers: 0.65 m/s2; 5.7°.
Choose the coordinate system with x-axis directed to the right and the y-axis directed downwards.
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
= mg cos θ and FT
= mg sin θ, where θ = 9.5°.
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
mg cos θ. The resultant of all forces has the magnitude F = FT
θ – μkcos
θ) and creates the acceleration a = F/m = g(sin θ – μkcos
(sin 9.5° - 0.10 * cos 9.5°) = 0.65 m/s2.
answer the second question we need to solve an equation a = 0.
sin θ – μkcos
θ = 0, tan θ = μk,
and θ = tan–1
0.10 = 5.7°.
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