Timeline for How is the work done to push a planet over 1m with 1N the same as pushing a feather over 1m with 1N?
Current License: CC BY-SA 3.0
10 events
| when toggle format | what | by | license | comment | |
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| May 23, 2017 at 17:13 | comment | added | Dedados | To apply the same force in an environment not affected by other forces or objects (vacuum with no gravity) they must work at the same rate to apply the same force, it doesn't make sense any other way. | |
| May 23, 2017 at 15:56 | comment | added | nasu | They don't have to work at the same rate. Actually for a rocket the efficiency depends on the speed so the power used to apply the same force may be quite different in the two cases. But this again, has nothing to do with energy conservation or work-energy theorem. You can even apply this 1 N forever with no work done on the planet (or feather). Just imagine something applying an 1N in opposite direction. | |
| May 23, 2017 at 15:07 | comment | added | Dedados | @JMac What you are saying makes sense but then I think that in order to produce the 1N of thrust, the rockets have to both work at the same rate, so how can they have a different power? The energy they output each second must be the same? | |
| May 23, 2017 at 14:57 | comment | added | JMac | @Dedados You gave the same amount of force in the same amount of distance. Assuming the rockets have the same efficiency in these setups, the time taken does not matter. One rocket would take more power (work per unit time), while the other one would have to apply the lower power for a longer time. | |
| May 23, 2017 at 14:56 | comment | added | nasu | So no, you don't know that the KE of the systems are the same in the end. Just that the planet and the feather have the same KE. | |
| May 23, 2017 at 14:55 | comment | added | nasu | You confuse "system" with just one part of it. The feather and the planet have the same KE in the end. So the work done on them is the same. This is all that work-energy theorem has to say about it. Think about a weight suspended from a hovering helicopter and the same weight sitting on the table. In both cases the work done on the weight is zero. But in the first case the helicopter uses a lot of fuel and accelerates lots of air (so increases its KE). In the second case there is no energy spent. The KE of the weight are the same in both cases but the final KE of the systems are not. | |
| May 23, 2017 at 14:44 | comment | added | Dedados | Since the KE in both systems is identical in the end after they have traveled the 1m with a force of 1N pushing them, the input energy must be the same for both of them and that equal to the KE since we are not taking into account inefficiencies, right? But, if the rockets use different amounts of fuel as they fire for different amounts of time, the input energy must be different right? So, how can this be? BTW, by rocket I mean rocket motor/engine, not space shuttle. | |
| May 23, 2017 at 14:36 | comment | added | nasu | Energy conservation in what system? You are comparing two systems. Energy conservation refers to comparing a system in two different states (initial and final). If you look at any of the two systems, for example rocket and planet, the energy at the beginning of the pushing and the energy after pushing 1m, the total energy is conserved. You should of course include all parts of the system (ejected gases too) and all forms of energy, not just KE. Conservation does not mean energy in different systems must be the same or has to be distributed in the same way. | |
| May 23, 2017 at 14:06 | comment | added | Dedados | The question was about how energy does not seem to be conserved, but yes, the KE must be the same for both of them. However, the input energy is different since the planet's rocket fires for a long time while the feather's for a very short while. Thanks for the answer though. | |
| May 23, 2017 at 13:50 | history | answered | nasu | CC BY-SA 3.0 |