No, it’s not. V1 and V2 in the expression you wrote are already deltas. So eccentialy what you are trying to represent as the change of KE is:
delta KE2 - delta KE1.
In other words you found how the gain of KE changed going from point 0 to point 2 compared to going from point 0 to point 1.
Ok, well then I didn’t know what situation you were envisioning. As I read your original post it seems like your V0 is the speed at which you are said to be stopped in the chosen reference frame. But if a 2kg object undergoes a 1m/s^2 constant acceleration and it goes from 1 m/s to 2 m/s in 1s that is a change in KE of 1 J to 4 J, and in the next second it will go from 2 m/s to 3 m/s which is a change of 4 J to 9 J. So I don’t see how you think the change in KE is the same in equal time intervals. Are you supposing you are in the rocket and your reference frame is accelerating with the rocket or something? I’m not following your logic.
In this case 1J is the gain of KE during the first second of going (between t0=0s and t1=1s), whereas 4J is the gain of KE during the first two seconds of going (between t0=0s and t2=2s. If you want to know how the KE changed going from 1 m/s to 2 m/s you have to take the differece in velocity between t1=1s and t2=2s, you will get 1m/s, so the change of KE between t1 and t2 would be 1J. The thing is that it’s nonsensical to talk about KE as having absolute nature, it’s by definition a relative one, you can only think about it’s change over a certain time interval.
I don’t see how 1 J in the first second plus 1 J in the second second makes your 4 J total. KE may be relative, but it’s relative to the reference frame, it seems like we’re agreeing with a (non-accelerating) reference frame that is stopped compared to the rocket at t=0. The work-kinetic energy theorem will show that the 2kg object having a force of 2N (creating the a = 1m/s^2) will act over distance d = 1/2 (vi + vf ) t = 1/2 (0 +1)(1) = 0.5m for the first second, so delta KE = Fnet d cos theta = (2N) (0.5 m) * cos 0 = 1 J. From t=1s to t=2s, d = 1/2 (1 + 2) (1) = 1.5m. So delta KE = (2N) (1.5m) cos 0 = 3J. So the change in KE from 1s to 2s is 3J (making a total of 4J at t=2s).
If you can find any respected physicist agreeing with you or some article from a respected source please share, but otherwise it seems like everything I know and everyone here disagrees with you.
You can not just sum up values defined in different frames. The gain of KE=1J in the first part of the motion is defined in stationary RF with respect to the KE=1J defined in the RF moving at 1m/s relative to the stationary one.
The definition of work (Wnet=Fnet(xf-xi)) implyes that the displacement (xf-xi), which is frame dependent, is only due to the force causing it.
In the reference frame stopped at t=0 the displacement is only due to the force applied. During the second part of the trip it is caused by the force** as well as the initial velocity**, which is 1m/s (at the beginning of the second part). So to define the work (W) exerted by the force (F) during the second part of the trip we should choose the reference frame stationary with respect to the object in the beginning of the second interval, i.g. moving at the speed 1m/s. In this reference frame the displacement d = 1/2 (0 + 1) (1) = 0.5m will only be due to the force and work would be 1J.
The results you get while playing with conservation of energy laws swithcing frames of reference may be very counterintuitive.
I believe physics is not so much about “reliable sources” or “respected scientists” as about understanding.
The best straightforward explanation, I could find, is here in the post #13.
Reference The final explanation to why kinetic energy is proportional to velocity squared | Physics Forums
Hmm… I’m not sure why then we are even allowed to keep the t=0 reference frame for 1s.
The results when switching reference frames might be counterintuitive, but I was not switching frames. It seems that you are and showing that if you keep switching the reference frame to the one at rest with the rocket then the gain in KE is the same each second. That is true, but not what the original discussion was about, which was the gain in KE each second when keeping the t=0 reference frame.
Since you seem to be saying that the change in KE is not greater each second, then it seems to me that you would not agree with the simple solutions to how fast an object would be going after dropping a height h, using conservation of mechanical energy.
Of course the universe it was it is, but if you are questioning reliable sources and respected scientists not on things being currently researched but things rather well accepted and used makes it sound like you think there’s a conspiracy theory. I did read some of the link you posted but I didn’t read all of it, and so far I’m not convinced of anything.
Hmm… I’m not sure why then we are even allowed to keep the t=0 reference frame for 1s.
The results when switching reference frames might be counterintuitive, but I was not switching frames. It seems that you are and showing that if you keep switching the reference frame to the one at rest with the rocket then the gain in KE is the same each second. That is true, but not what the original discussion was about, which was the gain in KE each second when keeping the t=0 reference frame.
Since you seem to be saying that the change in KE is not greater each second, then it seems to me that you would not agree with the simple solutions to how fast an object would be going after dropping a height h, using conservation of mechanical energy.
Of course the universe it was it is, but if you are questioning reliable sources and respected scientists not on things being currently researched but things rather well accepted and used makes it sound like you think there’s a conspiracy theory. I did read some of the link you posted but I didn’t read all of it, and so far I’m not convinced of anything. If you can explain it yourself then fine.
Right; momentum is conserved. If you simply dump something off the back, it will have lost no momentum, so the rocket will have gained none. If instead you energetically propel it backward, it will end up with less momentum, and the rocket will have more.
No, that would be ridiculous. I think I have to admit that throughout my posting I was thrown of by another puzzle and envisaged a different set up. The KE definately increases quadratically with time in the t=0 RF, which means for an earthbound vehicle the consumption rises each second. For a rocket moving in space the consumption is constant but acceleration rises each second due to decreasing mass. We can not ignore the changing mass in this case.
To be more exect, the ever decreasing mass of what is left of the rocket counterbalances the quadratic inrease of KE, making it linear.
If you move to using impulse, which is simply the delta of momentum it will be far less confusing than KE.
The question pertains to velocity and not how long it will take to stop or how much damage it will do in a collision. KE’s nature will confuse that question.
(assuming you stay well below 1/2c where you need to move to more modern physics anyway)