5 November, 11:01
a 62kg person is going down the hill slopes at 32°. the coefficient of kinetic friction between the skis and the snow is 0.15. how fast is the skier going 5.0s after starting from rest
5 November, 11:23
Draw a free body diagram of the skier. There are three forces: normal force pushing perpendicular to the slope, weight pulling down, and friction force pushing up the slope.
Sum of the forces perpendicular to the slope:
∑F = ma
N - mg cos θ = 0
N = mg cos θ
Sum of the forces parallel to the slope:
∑F = ma
mg sin θ - Nμ = ma
mg sin θ - (mg cos θ) μ = ma
g (sin θ - μ cos θ) = a
Given g = 9.8 m/s², θ = 32°, and μ = 0.15:
a = (9.8 m/s²) (sin 32° - (0.15) cos 32°)
a = 3.95 m/s²
After 5.0 s, the velocity is:
v = at + v₀
v = (3.95 m/s²) (5.0 s) + (0 m/s)
v = 19.7 m/s
Rounded to two significant figures, the skier's velocity is 20. m/s.
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» a 62kg person is going down the hill slopes at 32°. the coefficient of kinetic friction between the skis and the snow is 0.15. how fast is the skier going 5.0s after starting from rest