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Consider you are travelling on a board in the space moving with 5m/s velocity. There is no force acting on it. Now consider you have 2 stones of 1kg in your hand. So the stones also had 5m/s velocity.

Now you throw both stones in opposite directions with same force providing them with same kinetic energy (i.e one in front and other in back) with a velocity of 10m/s instantaneously. According to a person who is at rest and observing you will say that the stone you threw in front had a velocity of 15m/s and the stone you threw in back had a velocity of 5m/s.

Now from the stand point of energy conservation my question is the stone you threw in front had a initial velocity of 5m/s (i.e 12.5 joules of kinetic energy) you also did some work on it and increased its velocity (you gave it 100 joules of kinetic energy) so it has 112.5 joules of kinetic energy (i.e 15m/s velocity). But for the stone you threw on back side had initial 5m/s velocity (i.e 12.5 joules of kinetic energy) but you also did work to reverse its direction that is 100 joules of energy. But finally it has again 12.5 joules of energy. So where did this energy go?

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closed as unclear what you're asking by John Rennie, Jon Custer, glS, user191954, sammy gerbil Sep 26 '18 at 18:36

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This question is based on a false premise. Specifically “But for the stone you threw on back side had initial 5m/s velocity (i.e 12.5 joules of kinetic energy) but you also did work to reverse its direction that is 100 joules of energy.”

In fact, there was 0 work done on the stone thrown backwards. Remember,

$$W=\int_S F\cdot ds$$

During the first half of the push F and ds are in opposite directions, and so the work is negative. During the second half they are positive, and so the work is positive. The work in the two halves cancels out to 0 work overall.

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  • $\begingroup$ If you do no work on the 2nd stone it would have gained k.e. $\endgroup$ – ZeroTheHero Sep 22 '18 at 12:47
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This is not a bad question, it is actually a good question: but it is unfortunately tied to its particulars in a way that makes things difficult. You are right to sense a strange contradiction that is somehow "under the water."

Let's take a second to reveal it. The work-energy theorem says that the change in the kinetic energy of a particle moving with velocity $\vec v$ is given by the dot product of the net force with that velocity vector, which we call the net power exerted by those forces$$\frac{\mathrm dK}{\mathrm dt} = \frac{\mathrm d\hphantom{t}}{\mathrm dt}\left(\frac12 m\vec v\cdot \vec v\right) = \sum_i \vec F_i\cdot \vec v = P.$$The work-energy theorem in its usual form just allows this to happen for a certain unit of time: the power summed up over time gives the work $\vec F \cdot \Delta \vec r$ while the time-rate-of-change of kinetic energy summed up over time gives the change in kinetic energy $\Delta K.$

The problem you have noticed is that energy follows a conservation law, which means you can treat it as a sort of stuff. Like just as you might get frustrated, "I have 11 socks in front of me and I know I washed twelve, where is that twelfth sock" the laws of physics also justify a frustration whenever "I have 11 J of energy after a collision and I know I started with 12 J: where is that last joule?"

But the above equation doesn't act like any other stuff that you know, and why? Because of different reference frames. Normally when you shift to a reference frame moving with velocity $\vec u$, the amount of stuff in a box stays the same. But the kinetic energy of that box changes to $$K' =\frac12m\big(\vec v-\vec u\big)\cdot\big(\vec v-\vec u\big) = \frac12 m v^2 - m\vec u \cdot \vec v + \frac12 m u^2.$$ If it acted like any other stuff you know, there would be only that first term $K=\frac12 m v^2$. The second and third terms obviously don't cancel each other out unless either $\vec u = \vec 0$ and you didn't boost in the first place, or $2\vec v = \vec u$ and thus the particle has perfectly reversed its velocity during the shift. But those are obviously not general assumptions we can make!

You can also see this as a property of the power exerted, it was $P$ and now it must be $$P' = \sum_i\vec F_i\cdot\big(\vec v - \vec u) = P - \sum_i\vec F_i \cdot \vec u.$$

This is the part in an answer where I would say "and here's where you need to look at this from a different perspective," but in this case you actually need to keep looking from the same perspective. You are right that energy doesn't work that way. It is a pseudo-stuff, if you like: as long as you choose a reference frame and remain in it, you can treat it like it's a countable thing in the world. But it's not quite true because while energy conservation holds in two separate frames of reference, they do not agree on the numerical value of the energy that they are conserving. The problem with your understanding is that you're trying to calculate the work in one reference frame and use that as the energy change in another reference frame, and energy is not that convenient.

But note that all of the math works out just fine. The time derivative of a constant like $\frac12 m u^2$ is zero, and so we have $$\frac{\mathrm dK'}{\mathrm dt} = \frac{\mathrm dK}{\mathrm dt} - m\vec u \cdot \frac{\mathrm d\vec v}{\mathrm dt} = P - \vec u\cdot (m\vec a) = P - \vec u \cdot \sum_i \vec F_i = P'.$$ This is a long mathematical check on the statement we already knew: saying, "the work-energy theorem was derived from Newton's laws which are invariant under a change of reference frame."

In fact in many cases it works out better than fine. If you sum up forces on different bodies then due to Newton's third law they come in opposite-force pairs, and the terms $\vec u \cdot \vec F_i$ cancel due to having opposite signs. This means that this $-m\vec u \cdot \vec v$ term is eliminated from the sums, and so we just need to get rid of the $\frac12 m u^2$, which we can do by only looking at differences in the kinetic energy.

So if you were looking at two trains of the same mass and speed crashing into each other head-on and coming to rest, in the reference frame where they are both moving the same speed you would have, for either train, a change in kinetic energy from some $K$ to $0$. That energy of $(K - 0) + (K - 0) = 2K$ needs to be dissipated in the collision, for each of these trains.

But what about in the reference frame where the other one is at rest? Since it's a $v^2$ law and you are doubling velocity one of these must have started with $4K$ of energy! Is that a contradiction? No, but it takes a second to see why: afterwards both trains are still moving with speed $v$ in some direction so each one ends with $K$ energy. So the difference is instead $(4K - K) + (0 - K) = 2K.$ That energy difference has thus proven itself to be independent of reference frame, and the underlying reasons why are that (a) we are insisting on looking at differences in kinetic energy, and (b) the forces causing the energy loss are internal and therefore Newton's third law lets us cancel them.

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