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17 July, 14:11

A boy pulls a 28.0-kg box with a 230-N force at 35° above a horizontal surface. If the coefficient of kinetic friction between the box and the horizontal surface is 0.24 and the box is pulled a distance of 30.0 m, what is the work done by the friction force on the box?

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  1. 17 July, 14:38
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    1977.696 J

    Explanation:

    Given;

    Weight of the box = 28.0 kg

    Force applied by the boy = 230 N

    angle between the horizontal and the force = 35°

    Therefore,

    the horizontal component of the force = 230 * cosθ

    = 230 * cos 35°

    = 188.405 N

    Coefficient of kinetic friction, μ = 0.24

    Force by friction, f = μN

    here,

    N = Normal force = Mass * acceleration due to gravity

    or

    N = 28 * 9.81 = 274.68 N

    therefore,

    f = 0.24 * 274.68

    or

    f = 65.9232 N

    Now,

    work done by the boy, W₁ = 188.405 N * Displacement

    = 188.405 N * 30

    = 5652.15 J

    and,

    the

    work done by the friction, W₂ = - 65.9232 N * Displacement

    = - 65.9232 N * 30 m

    = - 1977.696 J

    [ since the friction force acts opposite to the direction of motion, therefore the workdone will be negative]
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