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# 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

- ?Lv 71 month agoFavorite 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

- oubaasLv 71 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