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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

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  1. 5 November, 11:23
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    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.
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