Same old question, I still don't get it: E=1/2mv^2 - more E with V?

[ryanabc67]

TL;DR Summary

Q: "does it take more energy to accelerate at higher speeds"
A: The 'usual answer': "Your kinetic energy grows exponentially"....yeah, but how much FUEL do you burn?

I still don't get this and most answers all over kind of miss the point and just kind of recite without really explaining.
Here's the dilemma. Yes, kinetic energy grows exponentially with speed. So a 1kg object changing from 0 to 1m/s increases by .5J. And from 4 to 5m/s increases from 8J to 12.5J = 4.5J.

But the question is NOT how much kinetic energy does a body have as it accelerates. The question is, how much energy does it REQUIRE to accelerate? This is not the same question.

Clearly, speed is totally relative and therefore, so is kinetic energy. If you are running towards me at 4m/s and your friend is stationary when I begin moving towards you both, my kinetic energy increase relative to you is 4.5J but only .5J relative to your friend, so, no, sorry, CHANGE IN KINETIC ENERGY DOES NOT EQUAL CHANGE IN REQUIRED ENERGY.

The question is about how much energy is required to accelerate and given the laws of "relativity" (Galilean), when I'm travelling 4m/s .... relative to me - I'm stationary. No matter how fast I'm travelling, there is no speed that can be assigned to me. Whatever speed 'you' think I'm travelling is just your own opinion and has nothing to do with me. Everyone will have a different opinion on that, depending on what direction they are moving and how fast.

So my speed is always zero and every increase I make is from 0 to 1 (let's assume I pause after each increase for a bit).

What doesn't make sense however, is the following scenario.
Caveat: I'm NOT arguing with the law of conservation of energy. Clearly THE FOLLOWING IS WRONG... I just don't understand why.

So, why is this wrong; what am I missing here...

I'm sitting in space with my jetpack on. I weigh 2kg (not 1kg). I'm stationary and not moving.
I look out and see the earth travelling towards me at 100m/s. And I see the moon right beside it travelling at me at 10m/s.
I'm going to crash into one of them because both the world and moon need more energy and they will absorb my kinetic energy. And believe it or not, they'll pay me for it. They'll pay me $1 for every Joule of energy when I crash.

But I'm kind of broke. It costs me $1 for every Joule of energy I expend, and I only have $2. But I figure, this is still a pretty good deal:
I will turn towards them and accelerate from 0m/s to 1m/s.
So for the earth my kinetic energy just went from 10,000 to 10,201J, and for the moon 100 to 121J.
Nice!!! it only cost me $1 and one joule of energy and in return I'll expend and extra 201J on Earth. I think I'll head towards Earth. That's $201 bucks in my pocket for just $1 cost.

So I figure it's just like with beer, if one is good then 2 must be better, so, I accelerate again.
But this time, what I see is the earth travelling at me at 101m/s. Sure 1m/s was from me, but space is relative, so, I'll just pretend I'm stationary again since no one can actually argue with that, and, so again I accelerate from 0 to 1m/s towards the Earth.
But this time my kinetic energy goes up to 10, 404J, or a whopping increase of 203J, or $203. Big party when I get out of the hospital.

So, yeah, I crash, go unconscious and wake up with pretty nurses all around. And I decide, hey, if this worked once, why not do it again?
So I spend $200 on a new jetpack and fly away from Earth back where I came from at 1m/s. That only costs me a buck.
(I'm travelling "with" the Earth here so I only need a little bit of speed. Also, we can ignore the cost of gravity since it turned out I crashed much harder on the way in because of gravity but that nullifies the cost of leaving against it).

So I wait a day or so until I'm good and far away, just coasting at a constant 1m/s. Then I turn around and try to do it all over again - but, hey wait, the Earth is only coming at me at -1m/s. This time I need to accelerate all the way up to about 100m/s. This was a bad idea and I learn my lesson about the law of conservation of energy.

But, this story strayed far from the normal explanations of this. And this leads to a case where the kinetic energy is just stored up in the universe but could all be used up if everything just crashed into everything else. What I want clarification on is the 1J turning into 203J and the idea that every 1m/s increase costs 1J here....which makes me wonder if instead I chose a span of .00001m/s to increase by instead. It seems maybe the kinetic energy formula has nothing at all to do with the cost of acceleration and how much energy it requires.

 

[russ_watters]

Long post, and I didn't read until the end, I stopped after the running scenario: frame dependent isn't the same as not conserved and changing the scenario in the middle of a problem will of course give wonky results. The answer is what the equation says at face value: accelerating to twice the speed takes 4x the energy.

 

[ryanabc67]

I suspect your response is that it says at face value; you didn't read it.

 

[jbriggs444]

I suspect your response is that it says at face value; you didn't read it.

Nonetheless, @russ_watters got it right. Energy is conserved but not invariant. You say that you understand this but you go on to demonstrate that you do not.

It is not possible to simply expend energy to change your velocity. You must transfer momentum as well. Whatever you are transferring momentum to will gain or lose energy as a result.

You can pick a reference frame where you are moving fast so that the energy gain you get from pushing off from a co-moving object is large. Yay for efficiency!!

But if you turn around and look at the result for that co-moving object, it will have lost energy. Boo for efficiency :-(

TANSTAAFL. You cannot get something for nothing.

 

[Ibix]

TL;DR Summary: Q: "does it take more energy to accelerate at higher speeds"
A: The 'usual answer': "Your kinetic energy grows exponentially"....yeah, but how much FUEL do you burn?

CHANGE IN KINETIC ENERGY DOES NOT EQUAL CHANGE IN REQUIRED ENERGY.

Yes it does. Remember that if you start running, the Earth changes speed too. Factor that energy change in too and you'll find your lost energy.

Consider two bodies, masses $m$ and $M$ that are initially at rest. They push off each other, acquiring velocities $v$ and $V$ respectively. Conservation of momentum tells you that $v=-MV/m$. What was the energy change? Since the initial KE was zero, it's just $\frac 12mv^2+\frac 12MV^2$.

Now analyse again, in a frame where the initial velocities were $u$. The final velocities are $V+u$ and $v+u$. What's the energy change? $$\begin{eqnarray*}
\Delta E&=&\frac 12m(v+u)^2+\frac 12M(V+u)^2-\frac 12mu^2-\frac 12Mu^2\\
&=&\frac 12mv^2+\frac 12MV^2+mvu+MVu\\
&=&\frac 12mv^2+\frac 12MV^2+m(-MV/m)u+MVu\\
&=&\frac 12mv^2+\frac 12MV^2
\end{eqnarray*}$$This is the same as before. But you will not get the same answer if you disregard the energy given to one object - that is where you are going wrong, as noted by both posters above.

 

[jack action]

So, why is this wrong; what am I missing here...

Every time you accelerate, you must lose mass.

I'm sitting in space with my jetpack on. I weigh 2kg (not 1kg). I'm stationary and not moving.

Once you begin accelerating, you won't be 2 kg of mass anymore. You will have to push some part of you (the fuel in the jet pack) in the opposite direction, forever lost to you, now going at a certain speed away from you.

When you are on Earth, your mass is your combined masses, not just 2 kg of you. You can push yourself off the Earth. The Earth - or part of it - will be the mass acquiring momentum away from you.

 

[Dale]

I just don't understand why.

It is wrong because you neglected the conservation of momentum.

As others have said, energy is conserved and it is also frame variant. But conservation of energy is not the only important conservation law. You must also consider conservation of momentum.

When you accelerate in your jet pack you will release a tremendous amount of energy. In a frame where you have a low velocity the vast majority of that energy will go into the exhaust. In earths frame more goes into you and less into the exhaust. You cannot neglect conservation of momentum in any frame and get the right answer

 

[russ_watters]

Every time you accelerate, you must lose mass.


Once you begin accelerating, you won't be 2 kg of mass anymore. You will have to push some part of you (the fuel in the jet pack) in the opposite direction, forever lost to you, now going at a certain speed away from you.

Pros and cons to different scenarios, but one reason I prefer cars to rockets here is there is little(gas) or no(electric) change in mass.

Also, others pointed out the Newton's 3rd Law implications but my gut tells me(without calculating) the "missing energy" is negligible for a car that acts against the Earth.

The bigger "missing energy" would be in comparing cars colliding head-on vs rear-ender at the same closing speed. That "missing energy" is resolved with conservation of momentum for an inelastic collission: the final speed of the cars in a rear-ender isn't zero.(As others have said)

 

[jbriggs444]

Also, others pointed out the Newton's 3rd Law implications but my gut tells me(without calculating) the "missing energy" is negligible for a car that acts against the Earth.

This is incorrect. It depends on what reference frame one adopts.

If one adopts a reference frame in which the road is moving then an accelerating car is doing non-negligible work on the moving Earth.

If you add it up, any unexpectedly high gain in kinetic energy because the road is moving is exactly compensated for by a loss in kinetic energy of the moving Earth.

 

[russ_watters]

This is incorrect. It depends on what reference frame one adopts.

If one adopts a reference frame in which the road is moving then an accelerating car is doing non-negligible work on the moving Earth.

I'm not sure what you mean by that; by "moving", do you mean accelerating while holding the car stationary?

 

[jbriggs444]

I'm not sure what you mean by that; by "moving", do you mean accelerating while holding the car stationary?

I mean adopting a reference frame where the Earth is moving.

For instance, one could adopt a reference frame in which the Sun is stationary and the Earth is moving at its orbital speed of some 107,000 kilometers per hour.

If a car is accelerating eastward at midnight (in the direction of the orbital motion), the force of the tires on the road will be draining energy from the Earth at a high rate. Power = force times velocity. The car gains kinetic energy rapidly in this scenario while the Earth loses kinetic energy rapidly.

 

[Dale]

Also, others pointed out the Newton's 3rd Law implications but my gut tells me(without calculating) the "missing energy" is negligible for a car that acts against the Earth.

It is only negligible in the inertial frame where the earth is initially at rest. In any other inertial frame it must be included.

 

[russ_watters]

I mean adopting a reference frame where the Earth is moving.

For instance, one could adopt a reference frame in which the Sun is stationary and the Earth is moving at its orbital speed of some 107,000 kilometers per hour.

If a car is accelerating eastward at midnight (in the direction of the orbital motion), the force of the tires on the road will be draining energy from the Earth at a high rate. Power = force times velocity. The car gains kinetic energy rapidly in this scenario while the Earth loses kinetic energy rapidly.

Fair enough, I didn't think of that.

Would you agree, though, that for a frame where they are at rest to start, the difference would be negligible or am I missing something there too?
[Edit: I see Dale confirms.]