Work on an incline

Question

   The diagram shows a crate with mass m=20.9kgm=20.9kg being pushed up an incline that makes an angle ϕ=24.7∘ϕ=24.7∘ with the horizontal. The pushing force is horizontal, with magnitude PP, and the coefficient of kinetic friction between the crate and the incline is μ=0.31μ=0.31. Consider the work done on the crate as it moves a distance d=4.33md=4.33m at constant speed.

 What is work, in joules, done by the pushing force, P?

Answer 1

Force is the integral of force with respect to position. The force is constant, so here, it is really just force times distance. Now, if the velocity is constant, what does that tell you about the net force?

--The net force should be zero, which means that the forces parallel to the incline should cancel out. So, we need to find the total force down the incline as the box is moving. If the angle of the incline is ϕ, you might recall that the weight of the box can be split into two components, mg sin(ϕ), which is parallel to the ramp, and mg cos(ϕ), which is perpendicular to the ramp.

The forces perpendicular to the ramp also cancel out, which means that the normal force N=mg cos(ϕ). You may recall that the friction force is the coefficient of kinetic friction multiplied by the normal force, and this moves in the opposite direction of motion. So the friction force is μN = μmg cos(ϕ).

So, given that the forces in the opposite direction are the component of gravity horizontal to the ramp (mg sin(ϕ)) and the friction (μmg cos(ϕ)), with what force is the box being pushed? (Remember that the velocity is constant). Then, what is the work? I hope this helps.