Energy Efficiency

I have no idea where to begin.

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To add to the advice given by Jaroc, here is a little more help using the diagram supplied.

Efficiency is the ratio of the work you do on the car (force times distance) to the (potential) energy it gains going up the slope.

More help is in the spoiler

Well, the work done in bringing the car up to the top would be :

FS + mg*S* sin alpha

I think you mean efficiency is useful energy obtained divided by word put in ??

Thus, mg S sinalpha / Fs

s cancels out leaving mg sin alpha / F ???

Well, that's not correct. As you can find it in Stonebridge's picture, work done is the scalar product of force and displacement (assuming that the force is constant). Here the force is parallel to displacement so work done is just $Fs$.

$mgs\sin \alpha$ is the change in potential energy of the car. It's the result of doing work over the car, not a part of the work done.

Edit: I have to disagree that efficiency is work done over energy gained by the car. It's the other way round. You put in some energy (doing work) and get 'useful energy' - the potential energy of the car.