A62kg 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

20. m/s

Explanation:

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

Substituting:

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.

0.19 m/s and 0.45 m/s