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.

143.815 N

Explanation:

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

component of gravity or mgsinθ friction or μmgcosθ

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