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13 February, 02:18

A rope pulls a 82.5 kg skier at a constant speed up a 18.7° slope with μk = 0.150. How much force does the rope exert?

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  1. 13 February, 03:15
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    374 N

    Explanation:

    N = normal force acting on the skier

    m = mass of the skier = 82.5

    From the force diagram, force equation perpendicular to the slope is given as

    N = mg Cos18.7

    μ = Coefficient of friction = 0.150

    frictional force is given as

    f = μN

    f = μmg Cos18.7

    F = force applied by the rope

    Force equation parallel to the slope is given as

    F - f - mg Sin18.7 = 0

    F - μmg Cos18.7 - mg Sin18.7 = 0

    F = μmg Cos18.7 + mg Sin18.7

    F = (0.150 x 82.5 x 9.8) Cos18.7 + (82.5 x 9.8) Sin18.7

    F = 374 N
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