### Where does the energy go?

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Geez... I'm sorry! :p
@LB: I see what you meant now. (Sounds foolish, but I... uh... er... skipped all those replies :'D )
 The British scientist and author C.P. Snow had an excellent way of remembering the three laws: First law: You cannot win (that is, you cannot get something for nothing, because matter and energy are conserved). Second law: You cannot break even (you cannot return to the same energy state, because there is always an increase in disorder; entropy always increases). Third law: You cannot get out of the game (because absolute zero is unattainable). http://www.physlink.com/education/askexperts/ae280.cfm

 An irreversible process increases the entropy of the universe. ... All complex natural processes are irreversible. ... During this transformation, there will be a certain amount of heat energy loss or dissipation due to intermolecular friction and collisions; energy that will not be recoverable if the process is reversed. http://en.wikipedia.org/wiki/Irreversible_process

ie. After every energy transformation, the amount of "free energy" decreases.
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Wait, a moment. Assume you are in a complete and perfect vacuum (i.e. a closed system), and there is a constant gravitational acceleration of 10m/s/s downwards. You have and object A on the ground (having energy only of its mass), and a single actor (an actor is any object that is capable of exerting force), that has enough energy to accelerate object A at 1m/s/s for 1s upward. Assuming the actor B the performs this action, the box will lift off the ground and at its highest point its gravitational potential energy would have reached its max (and thus it would be at its maximum energy).

The energy in the system at this point is accounted for because the actor now has no energy, and the object now stores this energy. The object will, due to gravity, fall downwards, loosing energy as it goes. Now assume we look at the system once the object has fallen back halfway, the object has lost half of the energy it gained when rising, but where has the energy gone?
Energy remains constant. ALWAYS.

Let's say our box has no energy and is resting. To raise it, we apply energy and it now haw potential: E = mgh.

When we drop the box, that potential is released by the box. It gets translated into kenetic energy as it falls E = 1/2mv2. When it hits the ground, the ground absorbs that impact.

Here's a good practical application: Solar panels provide power only when the sun is out. That may not correspond with electrical demand. Instead of having the solar panel power our house directly, we can pump water from a lower reservoir to a higher reservoir. Then when there is demand, we can let the water fall back down and drive turbines as it returns to the lower reservoir. This will regulate power and lets us store the solar energy as potential.
> Assume you are in a complete and perfect vacuum (i.e. a closed system)
Vacuum is space that is empty of matter.
a closed system can exchange energy (as heat or work) but not matter, with its surroundings.

I don't see how they are equivalent
Script Coder: Stewbond answered the question accurately but here's the simple rundown. When the box is resting, all energy is in the actor, the box has neither potential energy nor kinetic energy. When the box is atop the extended actor (its zenith), it is at max potential energy, 0 kinetic energy. All energy is currently in the box in the form of potential. When it drops it loses potential while simultaneously gaining kinetic energy. All energy is still in the box, but some of it is now kinetic. If soon after this point it where to hit ground, all energy (which would be in the box, and entirely kinetic) would be absorbed by that ground (which would deal with the kinetic energy it it's own manner). The kinetic energy absorbed by the ground would be equal the energy once stored within the actor.

EDIT: I may be wrong here but as a sidenote, would the kinetic energy halfway through the drop be a quarter of the total? Whilst the Potential energy retains the other three quarters?
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 Wait, a moment. Assume you are in a complete and perfect vacuum (i.e. a closed system), and there is a constant gravitational acceleration of 10m/s/s downwards. You have and object A on the ground (having energy only of its mass), and a single actor (an actor is any object that is capable of exerting force), that has enough energy to accelerate object A at 1m/s/s for 1s upward. Assuming the actor B the performs this action, the box will lift off the ground and at its highest point its gravitational potential energy would have reached its max (and thus it would be at its maximum energy).

If you apply an acceleration of 1ms-2 in the opposite direction of a constant accelerating force of 10ms-2, object A will not move as the resultant force will be 9ms-2 down.

 I may be wrong here but as a sidenote, would the kinetic energy halfway through the drop be a quarter of the total?

Consider an object B, with a mass of 100kg, in constant gravitational field of 10ms-2. We drop it from a height of 100m.

Initial gravitational potential energy of B: 100 * 10 * 100 = 100000 joules = 100 KJ
 ``` t = 0s Height: 100m Potential: 100KJ t = 1s Height: 90m Potential: 90KJ t = 2s Height: 70m Potential: 70KJ t = 3s Height: 40m Potential: 40KJ t = 4s Height: 0m Potential: 0KJ ```

Edit:
Actually, those figures are horribly approximate as it should be 4.5s total time to reach 0... I wonder if there is somewhere online you can make a nice graph of this?

I couldn't find somewhere to do a neat graph, but I did take the time to do the exact calculation (sorry if anyone sees any arithmetic errors, I was too lazy to use a calculator). I'm dropping object B, which weights 100kg, in gravitational field 10ms-2 from 125 meters (125m because it made the maths in my head easy. ;)).
 ``` Better figues: t = 0s Height: 125m Potential: 125KJ t = 1s Height: 120m Potential: 120KJ t = 2s Height: 105m Potential: 105KJ t = 3s Height: 80m Potential: 80KJ t = 4s Height: 55m Potential: 55KJ t = 5s Height: 0m Potential: 0KJ ```

(Useless fact of the day: I went a little further with the maths and figured the gravitational potential energy of an object with mass 100kg suspended about 1.1km high is about the same as the total energy you're going to use up today.) <-- That's all very bad arithmetic and don't listen to any of it.
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Mats wrote:
If you apply an acceleration of 1ms-2 in the opposite direction of a constant accelerating force of 10ms-2, object A will not move as the resultant force will be 9ms-2 down.

Me wrote:
single actor (an actor is any object that is capable of exerting force), that has enough energy to accelerate object A at 1m/s/s for 1s upward.

 Useless fact of the day: I went a little further with the maths and figured the gravitational potential energy of an object with mass 100kg suspended about 1.1km high is about the same as the total energy you're going to use up today.

Correct me if I am wrong, the object would have (assuming g=10m/s/s) (100*1100*10)/1000=1100JK of energy, right? I eat about 8000KJ of food a day, what happens to the other 7000KJ (btw, I am no where near fat, 49.6kg)?
 Correct me if I am wrong, the object would have (assuming g=10m/s/s) (100*1100*10)/1000=1100JK of energy, right? I eat about 8000KJ of food a day, what happens to the other 7000KJ (btw, I am no where near fat, 49.6kg)

Well spotted! I was working it out as the momentum on impact (1/2m v2), which of course will be equal to the maximum gravitational potential energy the object had (I was imagining that kind of energetic impact hitting the car outside)... And made a right f up of the calculation.

Something more like 400-450m would be more realistic. Anyway, my point was, that is a tremendous amount of energy. Imagine if that were a 100kg rock falling 400m. It hits your car at about 100ms-1 (over 200 mph). And we use up that much energy everyday. Crazy.

 single actor (an actor is any object that is capable of exerting force), that has enough energy to accelerate object A at 1m/s/s for 1s upward.

So here I guess you meant to accelerate object A at 11ms-2. Conventionally, you normally write gravity as - value and then other forces against gravity as + value to avoid confusions like that, but neither you or I bothered with convention in this case. :P
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Mats wrote:
Something more like 400-450m would be more realistic. Anyway, my point was, that is a tremendous amount of energy. Imagine if that were a 100kg rock falling 400m. It hits your car at about 100ms-1 (over 200 mph). And we use up that much energy everyday. Crazy.

Am I missing something? Because decreasing the height would surely decrease the potential energy (100 kg * 9.81 m s-2 * 500 m = ~4.91 kJ), and a 100 kg mass travelling at 100 m s-1 only uses about 500 kJ of energy. For a 100 kg object to have 2 000 kcal = ~8.3 MJ of potential energy, you would have to raise it to be about 8.5 km high. It would never have that much kinetic energy, though, because to do so it would need to attain a velocity of about 4 km s-1, while its terminal velocity would be at most 1 657 m -1 at s.t.p., assuming an area of 1 m2 and negligible drag.

 In case anyone is wondering, the difference between the potential and kinetic energies would be accounted for by e.g. heat generated by friction between the object and the air it's dropped in.
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Wow what is up with my maths today? Maybe I'll use a calculator...

 For a 100 kg object to have 2 000 kcal = ~8.3 MJ of potential energy, you would have to raise it to be about 8.5 km high

Thanks for the correct figure. That's even more insane!

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