Trying to non-technically explain why (kinetic) energy goes as $v^2$ rather then the perhaps-more-intuitive "double the speed, double the energy" $v$, I ended up putting my foot in my mouth (itself a difficult physical manipulation:) by "proving" precisely what I didn't want to prove, as follows.
Suppose you get an object moving to speed $v$. That'll require some amount of energy, which we'll just call $E_v$ (we won't need to bother calculating it). Now, to somebody else, moving alongside that object, it seems to be stationary. So he can get it moving at $v$ relative to himself with the same energy $E_v$ that you used to get it moving at $v$ relative to yourself.
So the total energy used is $2E_v$, first to get it moving from $0$ to $v$ relative to you, and then to get it moving from $0$ to $v$ relative to that other guy. But now it's moving at $2v$ relative to you, and the total energy used is $2E_v$ rather than the known-right answer $4E_v$.
So what's wrong? I'm guessing it's somehow got to involve those two different lab frames. For example, maybe whatever apparatus the second guy used in his frame first had to acquire it's $v$ relative to your original frame, and that required some energy. But even so, why exactly an extra $2E_v$ (to make up the "missing" difference)? So that can't be precisely the argument's error. But it's the two frames, somehow. Right? Or what?
>>Edit<< This is an extended reply to @StephenG's comment under @PhilipWood's answer below.Stephen: sure energy's a common-sense concept -- everything in physics is (must be) common sense if you can get to the underlying intuition. And after failing miserably with my above-described argument, I came up with a more successful attempt, described below just to prove my point that it ultimately must be common sense. This argument's a bit more elaborate, and I'd like to come up with a correct simpler one. But at least this argument gets the correct result...
Suppose you're hit with a ball going at speed $v$, and then with an identical ball going at speed $2v$. So how much "harder" does the $2v$ ball hit you?
To answer that, suppose the balls are made up of lots and lots of closely-packed identical little particles, all moving side-by-side together. Then each little $2v$-particle carries twice the "punch" of a $v$-particle ("punch" here is, of course, momentum, not energy, but I just said "punch" to avoid introducing big words and unnecessary technicalities).
However, since the $2v$-particles are travelling twice as fast, then in, say, one second, twice as many of them will hit you. Therefore, you'll be hit with twice as many particles, each carrying twice the "punch". So your "total punch" will be four times as great, not two times.
Okay, so this argument involves time, and therefore power rather than energy. So it's not entirely 100% achieving its purpose. But since this was a non-technical discussion, I simply didn't bother mentioning my misgivings about it. Good enough for the time being, I figured.
But, to elaborate my original question, can you make air-tight the above argument, and maybe explain what's wrong with the original one (hopefully so that it's correct and even simpler than this one)?