What is the minimum work needed to push a 950-kg car 710 m up along a $9.0 ^\circ$ incline? Ignore friction.

Hi lyonsca1,

Thanks for the question. I don't mean at all to contradict $W = Fd \cos(\theta)$ (which can also be written $W = F \cos(\theta) (d)$). That formula means that you multiply the force by the displacement, then multiply by cosine of the angle between them. In this particular solution, the force and displacement are actually parallel. The force I'm referring to is the force applied by some person or machine to make the car move up the ramp at constant speed. This applied force is directed along the ramp, so it's parallel to the length of the ramp. This means $W = F d \cos(0)$. Since $\cos(0) = 1$, it's common to not bother writing it at all, and to just write $W = F d$ for the special case where $\theta = 0$, which is the case where the force and displacement are parallel, as they are in this problem.

The applied force, according to the diagram, is $F_a = mg \sin(\theta)$, where theta here is referring to the angle of the incline. That's confusing, since the "theta" in the formula you wrote is referring to something different: namely the angle between the applied force and the displacement. If I were to write things out completely, using numbers in place of the angles, it would be $W = mg \sin(9.0^\circ) d \cos(0^\circ)$, where I added the $\cos(0^\circ)$ to try and be clear that that angle is the one in the $W = Fd\cos(\theta)$ formula, not the $9.0^\circ$ used to calculate the component of gravity along the incline.

Hope that helps,
Mr. Dychko