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I hope this is the right platform to ask stupid questions like this, but this is driving me nuts right now. Probably embarrassing to say, but I actually have somewhat of a physics background, but also definitely don't consider myself well educated in Physics. But I always thought my high school mechanical physics is somewhat solid. But this is causing me a major headache right now:

I'm well aware of $E_k = (1/2) mv^2$ but somehow never thought about it applied in the real world. Today it's been dawning on me that this equation is probably the reason why with a car it's so much easier to accelerate from 0-50 than from 50-100. I guess so far I always assumed that has to do with air resistance and the torque of the motor at different rpms and such. So first question: Is this really how kinetic energy works? Disregarding air resistance and other effects, it really takes 3 times as much energy to accelerate from 50 to 100 as it takes from 0 to 50?? Somehow that seems so counterintuitive to me.

And the more I think about it, the less it makes sense to me. Let's take an even simpler system: A rocket in space, pure vacuum, no interactions with anything. Let's say there is a reference point far away that is also not interacting with the rocket. How does it make sense that this rocket needs 3 times less fuel to accelerate from 0 to 50 as it does to accelerate from 50 to 100? Let's say, once I reach the speed of 50, I just change my frame of reference and consider the rocket to be at a speed of 0 again. If I then accelerate to 50 in the new reference frame (100 in the old) it surely can't use more fuel than for the initial acceleration.

Again, sorry for this probably really stupid question. I hope it's appropriate here. But this is really giving me a headache right now.

TLeksl
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  • This is a good question! When changing frames, account for the kinetic energy of the fuel which is expelled out of the end of the rocket. – DanDan0101 Apr 21 '23 at 17:05
  • Related: https://physics.stackexchange.com/q/535/2451 and links therein. – Qmechanic Apr 21 '23 at 17:15
  • I only understand half of what's said in those posted links. What I kind of get from what DanDan0101 posted is that in the rocket system ~^2 is misleading, and it does indeed take the same mass of fuel to accelerate from 0 to 50 as it takes from 50 to 100. Is that right? The thing with the kinetic energy seems to be that also the kinetic energy of the expelled fuel needs to be considered, so in the end it all adds up.

    But then how does that make sense with a case where chemical energy is converted into kinetic energy? Like in the case with of a car here on earth. I still don't get it!

     – TLeksl Apr 21 '23 at 17:28
  • With a car on earth, you have to account for the kinetic energy of the earth! – DanDan0101 Apr 21 '23 at 17:41
  • In a rocket chemical energy is converted into kinetic energy of both the rocket/payload and the kinetic energy of the exhaust. It's a very inefficient process. What upsets you is the inefficiency of it with respect to the payload kinetic energy, not the fundamental definition of kinetic energy. – FlatterMann Apr 21 '23 at 17:42
  • @DanDan0101 No, he doesn't. The car's engine transfers almost no kinetic energy to the planet. Almost all of it goes into the car. That's because the Earth has so much more mass than the car. Only if both had the same mass, then half the energy would end up in the car and half in the tiny Earth. – FlatterMann Apr 21 '23 at 17:43
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    @FlatterMann I mean, in a frame where the Earth is moving. – DanDan0101 Apr 21 '23 at 17:44
  • @DanDan0101 The only way to get to the kinetic energy in that frame is to have a collision with the planet. That's called "an asteroid impact". It's an easy way to get much more energy than even from nuclear weapons, indeed. – FlatterMann Apr 21 '23 at 17:46
  • @FlatterMann for clarification on what I mean, see the ending part of this answer – DanDan0101 Apr 21 '23 at 17:48
  • @FlatterMann DanDan is correct. In a frame where the earth is moving there can be a substantial transfer of KE to or from the earth. In general it must be accounted for – Dale Apr 21 '23 at 18:39
  • @TLeksl there are a large number of existing questions on this topic. Please search for them and read them and the answers. If they do not answer your question then edit this question to link to the other questions and explain in as much detail as you can what you need that is different – Dale Apr 21 '23 at 18:45
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    @FlatterMann in reference frames where the earth is moving it absolutely does transfer kinetic energy to or from the car – Dale Apr 21 '23 at 18:46
  • @FlatterMann see https://physics.stackexchange.com/questions/538885/why-does-this-example-seem-to-contradict-conservation-of-energy/538892#538892 for a general idea – Dale Apr 21 '23 at 19:34
  • Going back to the OP, why is it an issue that something takes 3x the energy to get from 50 to 100 than it takes to get from 0 to 50? That is simply correct, and expected from the maths. Have you not already tried to push something that is already moving? It gets more and more difficult to push! – naturallyInconsistent Apr 22 '23 at 15:09

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