Yahoo Answers is shutting down on May 4th, 2021 (Eastern Time) and the Yahoo Answers website is now in read-only mode. There will be no changes to other Yahoo properties or services, or your Yahoo account. You can find more information about the Yahoo Answers shutdown and how to download your data on this help page.

Anonymous
Anonymous asked in Science & MathematicsPhysics · 2 months 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

Attachment image

2 Answers

Relevance
  • ?
    Lv 7
    2 months ago
    Favorite 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

    Attachment image
  • oubaas
    Lv 7
    2 months 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

Still have questions? Get your answers by asking now.