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Anonymous asked in Science & MathematicsPhysics · 1 month ago

pushing box up an incline?

What force, F, is needed to push a 26-kg box at constant speed up an inclined plane that rises 5.0 m in a horizontal distance of 13 m if the coefficient of kinetic friction is 0.25? Include a free-body diagram as part of your solution

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2 Answers

  • NCS
    Lv 7
    1 month ago
    Favourite answer

    "constant speed" means equilibrium -- the sum of the downslope forces (component of weight + friction) is equal to the applied force:

    F = m*g*sinΘ + µ*m*g*cosΘ = m*g*(sinΘ + µ*cosΘ)

    Θ = arctan(5/13) = 21º

    F = 26kg * 9.8m/s² * (sin21º + 0.25*cos21º) = 151 N

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  • oubaas
    Lv 7
    1 month ago

    Θ = arctan 5/13 = 21.0 °

    sin Θ = 0.359

    cos Θ = 0.933

    F = m*g*(sin Θ+cos Θ*μ) = 26*9.806*(0.359+0.933*0.25) = 151 N

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