# Where does energy go when performing a useless effort?

I went to school one day, so I thought I was able to get this simple one.. but it looks like I'm not anymore. :(

One lonely little spaceship is resting into space. It has a small fuel capacity that it suddently burns to run away and finally reaches a speed $v$. Its tank is now empty, its kinetic energy $E = \frac{m v^2}{2}$ is exactly what it had in reserve, and it is equal to the work performed by the constant force $F$ it has just applied to itself over the run distance $d$ with $W = Fd = E$..

Now, instead of being free, it is tied with a rope to a wall. It does try to run away, burns its fuel.. But there is nothing to do, the rope is too strong. And it ends up with an empty tank and a null speed, a null kinetic energy, and the work $F$ has performed has been null all the time, for the rope's $-F$ has prevented any acceleration..

Where is its energy gone? The entire universe had energy $E$ all the time, the spaceship had $E$ initially (chemical potential) now $0$.. what is the system that had energy $0$ initially and now $E$, and why? what happened?

If it is only friction and heat within the rope and the wall, does it mean there is no way modeling a "perfect rope" without assuming that some energy is lost?

The answer is quite simple. You can't see it because you've forgotten that you make an unphysical simplification when considering "tied to an immovable wall" type situations. What's unphysical about the situation is quite simple : there's no such thing as an immovable wall. Let's say the rocket is tied to the earth. To say where the lost chemical potential energy has gone you would need to take into account the minuscule change in velocity that the earth has undergone. Because the mass of the earth is enormous the change in velocity required to produce the right amount of kinetic energy will be small, hence creating the impression of an "immovable wall".

As a side note, it's better not to think of things in terms of $W = Fd$ in this situation. It isn't wrong to do so, it's just harder (although not by much) to take into account the forces in play when the rocket is attached to a string. In particular, because the actual distance moved in such a situation is minuscule and unmeasurable for all intents and purposes, I don't consider it to be the most intuitive view of the problem. In addition, it's not trivial to define the distance moved when we are progressively excluding some of the fuel as being part of the system. All in all, to resolve your apparent paradox, it's simpler to think about things in terms of "which form of energy turns into which form of energy" for this situation, rather than consider the work done on a system, which is not trivial to define when the mass of the system is changing over time.

It should be noted that you've completely glossed over this "change of mass" subtlety in your analysis. The work done on the rocket is not the only thing involved. There's also work done on the gas as it ejects. As the rocket mass tends to $\infty$ (which is the equivalent of tying it to a very massive wall) most of the kinetic energy will indeed go to the gas, and you can't just ignore it.

• Plus, you don't get inelastic ropes. – Rob Jeffries Jul 27 '14 at 13:06
• @RobJeffries Indeed, that, and a number of other things we could mention don't exist in the real world. But even if you conducted a thought experiment where all those things did exist, there is no mechanism in classical physics that would let you imagine an immovable wall. That's why I focused on that, because that's where his missing energy comes from. – ticster Jul 27 '14 at 13:09
• Agreed. It's a bit like assuming you can "earth" a charged capacitor through a resistance-free circuit and then wonder where the energy has gone... – Rob Jeffries Jul 27 '14 at 13:15
• @RobJeffries Exactly, that's why I'm not focusing on the exact change in distribution of energy as the rocket mass increases (which is equivalent to attaching it) and focusing more on how he's created this situation by saying "Assuming this horribly unphysical scenario, why does this horribly unphysical result emerge from it?". – ticster Jul 27 '14 at 13:16

The rocket motor generates thrust by acelerating the exhaust gases that it emits. The force on the rocket is equal to the change in momentum of the exhaust gases per second, i.e. the exhaust velocity times the exhaust mass per second.

For the rocket travelling in space the energy generated from burning the fuel goes partly into the kinetic energy of the rocket and partly into the kinetic energy of the exhaust gases.

In your second example where the rocket is fixed in position, because the rocket can't move all the energy goes into the kinetic energy of the exhaust gases. Were the energy ends up is going to depend on what happens to those exhaust gases. Some of the exhaust will hit the wall and heat it up, so some of the energy goes into heating the wall. The rest probably just swirls around in the atmosphere and eventually slows to a stop. It's energy will go into heating up the atmosphere.

• "because the rocket can't move all the energy goes into the kinetic energy of the exhaust gases" The is just false. A part of the energy goes into setting into motion, however little, the heavy object to which the rocket is attached. – ticster Jul 27 '14 at 13:00
• While in the limit of heavy objects it's true that most of the energy will go into the gas, the reasoning used here is false and fails to address what the OP has misunderstood. Energy can go into the rocket/wall system, and the OP is simply failing to take into account all sources of energy. – ticster Jul 27 '14 at 13:42
• @ticster: yes, the wall is attached to the Earth, and the rocket is in principle moving the Earth slightly. However to include that in the answer seemed unnecessarily pedantic. – John Rennie Jul 27 '14 at 16:07
• @JohnRennie: or maybe necessarily, because it turns out that that's what I actually missed. ;) Suppose the rocket doesn't work with fuel and ejected gas, but with a bunch of healthy volunteers mounted on bicycles.. Then they would have ended up tired and they would --- just like me --- have wondered where the energy from their sandwiches had gone (certainly not in twirling gases outside :P). – iago-lito Jul 27 '14 at 20:08
• @ticster: (Note that in this case, the mass of the rocket wouldn't change over time, and so this looks rather like what I had in mind.) – iago-lito Jul 27 '14 at 20:08