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2 January, 13:21

A 65 Kg skier starts at rest at the top of a 150 m long hill that has an incline of 28 degrees. How fast will she be going at the bottom of there is 50N of friction?

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  1. 2 January, 13:34
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    34 m/s

    Explanation:

    Potential energy at top = kinetic energy at bottom + work done by friction

    PE = KE + W

    mgh = ½ mv² + Fd

    mg (d sin θ) = ½ mv² + Fd

    Solving for v:

    ½ mv² = mg (d sin θ) - Fd

    mv² = 2mg (d sin θ) - 2Fd

    v² = 2g (d sin θ) - 2Fd/m

    v = √ (2g (d sin θ) - 2Fd/m)

    Given g = 9.8 m/s², d = 150 m, θ = 28°, F = 50 N, and m = 65 kg:

    v = √ (2 (9.8 m/s²) (150 m sin 28°) - 2 (50 N) (150 m) / (65 kg))

    v = 33.9 m/s

    Rounded to two significant figures, her velocity at the bottom of the hill is 34 m/s.
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