Timeline for What conditions are required for the derivative of kinetic energy to be F.v?
Current License: CC BY-SA 3.0
11 events
| when toggle format | what | by | license | comment | |
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| Aug 2, 2017 at 18:53 | vote | accept | Daniel Underwood | ||
| Aug 2, 2017 at 18:52 | history | edited | Daniel Underwood | CC BY-SA 3.0 |
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| Aug 2, 2017 at 2:41 | history | edited | Daniel Underwood | CC BY-SA 3.0 |
Add alternate derivation by definition of work
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| Jul 31, 2017 at 16:11 | comment | added | Daniel Underwood | @mmesser314 Could that also be argued by forces perpendicular to velocity doing no work and the work-energy theorem? | |
| Jul 31, 2017 at 10:49 | answer | added | Farcher | timeline score: 2 | |
| Jul 31, 2017 at 6:25 | comment | added | mmesser314 | If the parallel component is in the direction of F, the force increases the speed, and hence increases T. If they are opposed, the force decreases the speed and T. The perpendicular component does neither. It changes only the direction of v. Think of an electron in a uniform B field, where F = qv x B. You will need to show this. | |
| Jul 31, 2017 at 4:41 | answer | added | CR Drost | timeline score: 0 | |
| Jul 31, 2017 at 3:51 | comment | added | Daniel Underwood | @mmesser314 it should change it in the same way that the parallel component would due to the square, correct? | |
| Jul 31, 2017 at 3:45 | history | edited | Daniel Underwood | CC BY-SA 3.0 |
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| Jul 31, 2017 at 3:24 | comment | added | mmesser314 | How much does the v perp component change T? | |
| Jul 31, 2017 at 1:47 | history | asked | Daniel Underwood | CC BY-SA 3.0 |