What force must be applied to push a carton weighing 245 n up a 15.0° incline, if the coefficient of kinetic friction is 0.340? assume the force is applied parallel to the incline and the velocity is constant.

Explanation:

when a person is pushing a carton in an inclined plane there is two forces acting against  the applied force,

the carton goes up ( constant velocity ) only if sum of these two forces becomes equal to applied force

from the question,

weight of the carton,mg = 245 N

angle of incline,θ = 15°

coefficient of friction, μ = 0.340

∵ the applied force, F = mgsinθ + μmgcosθ

                                    = mg ( sinθ + μcosθ)

                                    = 245 ( sin 15° + 0.340 cos 15°)

                                    = 245 ( 0.259 + 0. 328)

                                    = 143.815 N

0.11 m/s and 0.28 m/s should be correct