Timeline for Kinetic energy in different reference frames
Current License: CC BY-SA 3.0
13 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Jun 11, 2020 at 9:33 | history | edited | CommunityBot |
Commonmark migration
|
|
| Sep 30, 2017 at 9:57 | vote | accept | Markus Zetto | ||
| Sep 30, 2017 at 9:51 | history | edited | Kamil Maciorowski | CC BY-SA 3.0 |
added 15 characters in body
|
| Sep 30, 2017 at 9:43 | history | edited | Kamil Maciorowski | CC BY-SA 3.0 |
added 246 characters in body
|
| Sep 29, 2017 at 23:57 | history | edited | Kamil Maciorowski | CC BY-SA 3.0 |
added 266 characters in body
|
| Sep 29, 2017 at 23:45 | comment | added | Kamil Maciorowski | @Intergalakti There's some math now in my expanded answer. | |
| Sep 29, 2017 at 23:39 | history | edited | Kamil Maciorowski | CC BY-SA 3.0 |
elaborated answer
|
| Sep 29, 2017 at 15:16 | comment | added | Markus Zetto | But inserted in the formula above, $$v_1=\sqrt{\frac1m(2E_0+wmv_0)+w^2}-w$$ Is there something else I miss or is my calculation just wrong? | |
| Sep 29, 2017 at 15:09 | comment | added | Markus Zetto | I tried to include this effect in my calculation, but I still can't make it work out. First, on may recognize that the car moves only horizontally, so only the horizontal force will matter. Furthermore, $\Delta p = \int F dt$ and $$E=\int F ds = \int F \frac{ds}{dt} dt=w\Delta p=wm v_0 $$ | |
| Sep 29, 2017 at 13:21 | comment | added | garyp | This is an interesting and instructive question and answer. If you can spare the time to do it, I think we could benefit from a detailed analysis. | |
| Sep 29, 2017 at 13:17 | history | edited | ACuriousMind♦ | CC BY-SA 3.0 |
added 14 characters in body
|
| Sep 29, 2017 at 12:42 | history | edited | Kamil Maciorowski | CC BY-SA 3.0 |
added 4 characters in body
|
| Sep 29, 2017 at 12:34 | history | answered | Kamil Maciorowski | CC BY-SA 3.0 |