A 20.0-kg crate sits at rest at the bottom of a 11.0 m -long ramp that is inclined at 36 ∘ above the horizontal. A constant horizontal force of 290 N is applied to the crate to push it up the ramp. While the crate is moving, the ramp exerts a constant frictional force on it that has magnitude 65.0 N.

Question

The relevant forces in the direction of displacement:
Fx = Fcos(θ)-friction
Fy = mgsin(θ)

Putting it all together:
Work = (Fcos(θ) - friction - mgsin(θ))(11)

I know which forces to prioritize, but what I don't quite understand is why I have to subtract Fx-Fy when I plug in F to the Work equation.

Any clarification would be greatly appreciated.

Answer 1

Fx = Fcos(θ)-friction friction is not in the x direction. It is down the plane.
Work=force*distance (both vectors, the dot product). I wlll do this as dot products, with i and j vectors
= 65i(cosTheta(11icosTheta+11jsinTheta)+ -forcefriction(-i)*11iCostheta
well the second term -i*i is -1; the first tem i*j has a dot product of 1, and the second i*j is zero. you put the -force friction because you are pushing against it. force opposing friction=-forcefriction
so finally,
work= 65*11*cos²Theta + forcefriction*11CosTheta where the force opposing friction is 65*cosTheta