# Thread: What is the Potential Enenrgy of my vehicle?

1. Ok, so I've been having this argument with a friend of mine. If my vehicle is on a straight, flat piece of the world, what would be its potential energy? I say it would be 'x' . Any value larger than zero. Him, on the other side, say it would be zero. His reasoning for that answer is that Potential Energy = m (mass) x h (height) x g (gravity). In that equation, he is correct. I do agree with that. But to me, that equation would be for gravitational potential energy. My reasoning for it being larger than zero potential energy would be because of the fuel stored in the tank. And also the fact that you have electrical potential energy stored in the battery. The way I understand it is that the total energy before a given moment in time, must equal the total energy plus the losses after that moment in time. (Given the law of conservation of energy). And also, as far as I understand, the total potential energy of a system is the sum of all possible potential energies. Please can someone explain where I am wrong? And I say that I am wrong is because I have been proven to be wrong before. Thanks.

2.

3. Correct: the equation involving height is for gravitational potential energy.
Also correct: you have potential (chemical) energy in your fuel tank and (electrical) in the battery.

Ergo the potential energy of your car is x.
You're not wrong.

(Wiki IS a good place to look).

4. Yes, the fuel and the battery do count, but you're correct in saying he's referring to gravitational potential energy- in which case, he's correct for the gravitational potential energy.
For the total energy of the system, you're correct to say it's non-zero.

ETA: Jinx? I just refreshed the page and saw no replies. So I typed a reply, hit submit and saw Dwyddyr's mug staring back at me. Stupid forum.

5. the only problem is measuring all that accurately at any point in time. but if you were to do that why not also find the mass and use E=mc^2 just incase fission occurs and why not the fact that your car is already moving in order to keep a stationary position on the earths surface.

6. Originally Posted by fiveworlds
the only problem is measuring all that accurately at any point in time. but if you were to do that why not also find the mass and use E=mc^2 just incase fission occurs

7. Just to be thoroughly picky, I'd claim that the gravitational PE isn't zero, except with respect to the part of the road it's currently stood on.
It has GPE with respect to any elevation lower than than that point, by definition.

8. Originally Posted by Dywyddyr
Just to be thoroughly picky, I'd claim that the gravitational PE isn't zero, except with respect to the road it's currently stood on.
It has GPE with respect to any elevation lower than the road, by definition.
If you want to get that picky, you can claim non-zero due to bolted components within the vehicle that are at a greater height than where the tires meet the road.

9. True. I do agree with all. And wiki is a very good place for info, however searching on that, and any other site that I have been to, just gives the Potential Energy = mxgxh equation. That is what made me believe that I was wrong.

10. There is also an equation for electrical potential energy...

11. Uh, I just checked - the Wiki entry does mention elastic, chemical, electrical and nuclear PE.
No equation would be given for chemical since that depends on the chemicals involved - too many to list on that page.

12. Potential energy is the difference of energy between one state and another state that is considered as "the" reference (= 0 potential energy).

It is similar to the temperature, you can set the zero value differently for Kelvin, Celsius or Fahrenheit scales.

So, there is no absolute value for potential energy ; you must first decide the state that will be zero : in your example, you may decide that zero correspond to the car at sea level, battery and tank empty, the temperature of the car 0° Celsius and the car not being desintegrated ! But you may choose something else.

13. Yes. I get the reference point being an influential part of the equation. I have not argued that point anywhere. POR (point of reference) will always be there in any equation. But, to be able to use that in the equation, one must first consider complete standstill. No movement at all. Then you can use reference points (whether in space, geological or any other reference point you want to use) which will influence the outcome. However, I don't think it will change the object of having potential energy at 'stand still'.

14. Originally Posted by Dywyddyr
Uh, I just checked - the Wiki entry does mention elastic, chemical, electrical and nuclear PE.
No equation would be given for chemical since that depends on the chemicals involved - too many to list on that page.
I didn't say it was on the wiki page- I haven't looked at it. I was telling the O.P. another thing to google up. As in, "There's this..." and have him search for it and see what he finds.

15. My reply was to the post by WaterWalker (Peter Hamilton fan?) the one above yours, not yours. You just happened to get your post in before mine!

16. Originally Posted by Dywyddyr
My reply was to the post by WaterWalker (Peter Hamilton fan?) the one above yours, not yours. You just happened to get your post in before mine!
Yeah, well, my reply was to you, which was just after the one above and but second below the two above it. It referred to the one at that position, not the other position or any other position, for that matter.
It's the one above this one, but not one above this one as it's two above this one.
It's the twelfth post in the thread, and the one I replied to was the eleventh. It is not the thirteenth nor is it the tenth. It's more than ten but less than fourteen and it is to that exact one to which I replied.

I think.

17. How about this dilemma? Work = Force X Distance
Thus, if we use our physical ability to push, let's say, a vehicle, and we exert a constant 50 lbs. of force against it's bumper, while it moves 2 feet, the work we have done is 100 Lbs.-ft. Is this correct?

Now, let's assume the owner has locked his vehicle in a stationary position using the parking brake, and we apply our 50 lbs. of force, the distance moved is 0 ft., thus the work done is zero. Is this correct? But, if we apply our 50 lbs. for, say, 15 minutes, our muscles will cry out that they sure as hell did some work, will they not?

jocular

18. in this case would it not be similar to trying to push a wall? so force applied to the wall is similar to pushing yourself away with minimal wall movement. probably because the amount of kinetic energy required to start movement is a lot higher.

19. An interesting consideration! A more extreme comparison would be to stand with one's feet together, they being about 3 feet away from the wall, hands at the sides, and forehead pressed against the wall, supporting the body weight at an angle. Guaranteed to make one think he has done "work"! jocular

20. Ok. A bit off topic, but all quite valid. Potential energy does have a function for work done. As in the case of moving from point "A" to point "B" and back to point "A". I would like to thank all of you in your attempts to solve our little debate, but it has now escalated. Now, apparently, I'm using the wrong formuleas. Oh well. I give up.

21. The more correct equation for gravitational potential energy (from the earth) is,

where is the gravity constant, is the mass of the earth, is the mass of your car, and is the distance from the center of the earth to your car. As others have mentioned, with respect to the ground you can approximate your gravitational potential energy to be nearly zero. But if a chunk of the earth were removed beneath you the car would plummet to the bottom, releasing its gravitational potential energy.

22. Originally Posted by beefpatty
But if a chunk of the earth were removed beneath you the car would plummet to the bottom, releasing its gravitational potential energy.
I really like that word. Releasing. That, to me, states that the energy required for the car to fall already needs to be there. Ergo, potential energy.

23. Originally Posted by WaterWalker
I really like that word. Releasing. That, to me, states that the energy required for the car to fall already needs to be there. Ergo, potential energy.
I'm glad you caught that as I intentionally worded it that way . Although, it would be more accurate to say it is converted​ to kinetic energy.

24. Originally Posted by fiveworlds
in this case would it not be similar to trying to push a wall? so force applied to the wall is similar to pushing yourself away with minimal wall movement. probably because the amount of kinetic energy required to start movement is a lot higher.
No, when you push against a wall without moving it, you are not doing any work. If you are sitting in a chair it is pressing down on the floor with a force equal to your weight. No work is being done.

25. Originally Posted by WaterWalker
Ok. A bit off topic, but all quite valid. Potential energy does have a function for work done. As in the case of moving from point "A" to point "B" and back to point "A". I would like to thank all of you in your attempts to solve our little debate, but it has now escalated. Now, apparently, I'm using the wrong formuleas. Oh well. I give up.
What wrong formulas do you think you are using? If you move from point a to b and back to a, there isn't any net work done, unless friction is involved.

26. @beefpatty - True. And then you can even expand your formulations and equations further from that to include friction, sound, heat and various other sources of "lost" (or to be precise - converted) energy.

@Harold14370 - Yeah. That's the thing. I don't think the other guy can quite understand it when I try to tell him something, and to explain it with a "perfect" system. He just likes to overcomplicate things. Need to devise an experiment to test that theory.

27. Originally Posted by beefpatty
The more correct equation for gravitational potential energy (from the earth) is,

where is the gravity constant, is the mass of the earth, is the mass of your car, and is the distance from the center of the earth to your car. As others have mentioned, with respect to the ground you can approximate your gravitational potential energy to be nearly zero. But if a chunk of the earth were removed beneath you the car would plummet to the bottom, releasing its gravitational potential energy.
You left out the minus sign on the right half of the equation.

28. Thanks for pointing that out .

29. Gravitational effects are mutual; that is, an object dropped from some height falls to the ground, while exerting gravitational force of it's own, upon the earth. Thus, BOTH are actually in motion, driven towards one-another. Obviously, however, the big sphere only moves an infinitesimal amount upwards toward the falling object.

An interesting conjecture might be this: If the falling object were not originally a part of the earth's mass, that is, the object came from some other mass system, does the added mass of the combined two, after the object is at rest on the earth's surface, cause the earth's orbit to change slightly?

(what nonsense!) jocular

30. Originally Posted by Dywyddyr
Correct: the equation involving height is for gravitational potential energy.
Also correct: you have potential (chemical) energy in your fuel tank and (electrical) in the battery. ...
And E=m*c^2 is also potential energy.

31. Originally Posted by kowalskil
And E=m*c^2 is also potential energy.
While that is, strictly speaking, correct, it's not usual practise to convert anyone's car into energy.
The insurance companies tend to increase the premium. Generally for the owner and everyone else within a few city blocks.

32. Originally Posted by Dywyddyr
Originally Posted by kowalskil
And E=m*c^2 is also potential energy.
While that is, strictly speaking, correct, it's not usual practise to convert anyone's car into energy.
The insurance companies tend to increase the premium. Generally for the owner and everyone else within a few city blocks.
Wait, hold up. is this equation really for potential energy alone? Or does it include different types of the energies? (Kinetic, elastic, electrical etc...)
I always thought that because it has the velocity variable, it would be for kinetic energy?

33. Originally Posted by WaterWalker
Originally Posted by Dywyddyr
Originally Posted by kowalskil
And E=m*c^2 is also potential energy.
Originally Posted by WaterWalker
Originally Posted by Dywyddyr
While that is, strictly speaking, correct, it's not usual practise to convert anyone's car into energy.
The insurance companies tend to increase the premium. Generally for the owner and everyone else within a few city blocks.

Wait, hold up. is this equation really for potential energy alone? Or does it include different types of the energies? (Kinetic, elastic, electrical etc...)
I always thought that because it has the velocity variable, it would be for kinetic energy?

It includes every kind of potential energy, if you consider that the whole mass of the car can be converted into pure energy.

34. Originally Posted by WaterWalker
Wait, hold up. is this equation really for potential energy alone? Or does it include different types of the energies? (Kinetic, elastic, electrical etc...)
I always thought that because it has the velocity variable, it would be for kinetic energy?
It's not a velocity variable - it's a fixed constant, the speed of light in a vacuum.
And no, it doesn't take into account elastic, kinetic or any other type of energy: it's the direct conversion of the mass (regardless of condition/ state) into energy.

35. It is also possible to express it as a relativistic energy that takes into account the kinetic term if you cast the equation in the following form:

where is the invariant mass, or rest mass. But, it's usually just written as

.

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