Yahoo Answers is shutting down on 4 May 2021 (Eastern Time) and, as of 20 April 2021 (Eastern Time), the Yahoo Answers website will be 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 · 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 ### 2 Answers

Relevance
• 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 • 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 answers by asking now.