Well, it seems that after many discussions, we could agree in some topics about the question. I just copy and paste the answer. Two things. First, you can see that you only use friction to assume contact with earth (transfer of energy) but it didnt appear in the mathematical demostration, and you dont need to mention it. Second, the mention of battery is just illustrative, to remark that you have a difference of energy spent if you measure in different reference frame. The real thing you need to know is that kinetic energy is relative to the reference frame you use. *-*-*-*-*-*-*-*-*-*-*-*-*-*-*-*-*-*-*-*-*-*-*-*-*-*-*-*-* A force is acting between ground and the toy car. That is relevant. Momentum of Earth changes in this problem as well, and therefore, so does its energy. The extra 6 units of energy in above come from kinetic energy of Earth relative to the observer moving at speed of 3. Lets make this explicit, shall we? Mass of the car in the above problem must be m=2 to make numbers work. Lets take Earths mass to be M>>m. Ill take a limit where M->∞, because that will give me the correct answer. In Earths rest frame: Cars initial velocity: 0. Earths initial velocity: 0. Cars final velocity: 1. Earths final velocity: -m/M. Total final energy: E(M)=1+1/2M(−m/M)^2 lim (M→∞) E(M)=1 This is no different than your initial answer. Now, lets do this relative to observer heading towards the car at 3. Cars initial velocity: 3. Earths initial velocity: 3. Cars final velocity: 4. Earths final velocity: 3-m/M. Ei(M)=9+1/2M(3)^2 Ef(M)=16+1/2M(3−m/M)^2 E(M)=Ef(M)−Ei(M)=7+1/2M(−m/M)^2−3M (m/M) E(M)=1+1/2M(−m/M)^2 lim(M→∞)E(M)=1 Notice that even before the limit is taken, you get exactly the same result. In the limit, you get the same result as if you ignored the Earth in Earths rest frame. P.S. Yes, Im ignoring rotation of Earth for simplicity, but the logic is exactly the same. physicsforums/showthread.php?t=464859
Posted on: Thu, 31 Oct 2013 15:39:20 +0000
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