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23 June, 12:32

A ski starts from rest and slides down a 22 incline 75 m long. if the coefficient of friction is 0.090, what is the ski's speed at the base of the incline?

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  1. 23 June, 15:00
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    L = length of the incline = 75 m

    θ = angle of incline = 22 deg

    h = height of skier at the top of incline = L Sinθ = (75) Sin22 = 28.1 m

    μ = Coefficient of friction = 0.090

    N = normal force by the surface of incline

    mg Cosθ = Component of weight of skier normal to the surface of incline opposite to normal force N

    normal force "N" balances the component of weight opposite to it hence we get

    N = mg Cosθ

    frictional force acting on the skier is given as

    f = μN

    f = μmg Cosθ

    v = speed of skier at the bottom of incline

    Using conservation of energy

    potential energy at the top of incline = kinetic energy at the bottom + work done by frictional force

    mgh = f L + (0.5) m v²

    mgh = μmg Cosθ L + (0.5) m v²

    gh = μg Cosθ L + (0.5) v²

    (9.8 x 28.1) = (0.09 x 9.8 x 75) Cos22 + (0.5) v²

    v = 20.7 m/s
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