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# Energy Efficiency watch

1. I have no idea where to begin.
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3. (Original post by Ari Ben Canaan)
I have no idea where to begin.
In this question there is a silent assumption that the change kinetic energy of the car is 0 (this can be achieved in two ways - be either introducing resistive forces, or by setting the value of force F to be just sufficient to let the car be in equilibrium - and no greater).

Efficiency is the ratio of change in energy and work done. So, how did energy of the system change when the car was pulled onto the top (as I said, assume the change in kinetic energy of the car to be zero throughout movement)? How much work was done by the force?

The question is, why should you assume the change in kinetic energy of the car to be zero throughout the process? Because they apparently want you to do so. In fact, they didn't say that in the question so you could argue that efficiency is 100% - if there was no friction, no energy would be lost - or any other value - because you in fact don't know how much energy would be lost due to resistance. But to get the right answer you have to assume that the change in kinetic energy of the car is 0 because they assumed so writing this question - there's really no better reason with the information given.

Edit: Oops, sorry - I didn't notice the information that velocity was constant. So now we do have the reason to say that the change in kinetic energy is 0 - since the speed didn't change
4. 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

Spoiler:
Show

5. (Original post by jaroc)
x
Well, the work done in bringing the car up to the top would be :

FS + mg*S* sin alpha
6. (Original post by Stonebridge)
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

Spoiler:
Show

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 ???
7. (Original post by Ari Ben Canaan)
Well, the work done in bringing the car up to the top would be :

FS + mg*S* sin alpha
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 .

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.
8. (Original post by Ari Ben Canaan)
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 ???
Yes, I mean (meant) it is
potential energy obtained / work done
I've corrected the diagram.
9. (Original post by jaroc)
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 .

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.
Yes of course. I was upside down when I wrote that.
The answer is (energy obtained) / (work done) and not the other way round.

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