Timeline for Why is the potential energy of a spring the same when it is compressed and stretched?
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| Sep 10, 2017 at 8:55 | comment | added | Meni Rosenfeld | Agreed with @jwg. I think you are doing your students a disservice with your approach. Spring energy is quadratic because increasing $d$ increases both the average force and the distance. Better to teach them that, than to instill the notion that physics is just a bunch of random, unrelated concepts where you just make stuff up as you go along. I'm not sure why you would even want to discuss a spring's energy if the students don't even know the most basic thing about a spring's force. | |
| Sep 8, 2017 at 16:06 | answer | added | Jahan Claes | timeline score: 0 | |
| Sep 8, 2017 at 15:19 | answer | added | user16035 | timeline score: 0 | |
| Sep 8, 2017 at 9:15 | comment | added | jwg | This is the XY problem. You have started justifying that formula using several bad ideas instead of the one correct idea, Hooke's law. Now you want to help to demonstrate one of these steps, which isn't necessarily true. This is not going to eventually lead to an adequate answer. Your real problem is to understand and explain Hooke's law. | |
| Sep 7, 2017 at 23:36 | comment | added | alephzero | "But if a student argues that k could be defined with other units...", sorry, but I can't see any mathematics at all, from that point on. I can see a lot of pseudo-mathematical nonsense, though. | |
| Sep 7, 2017 at 23:28 | comment | added | alephzero | This doesn't even depend on the linearity aspect of Hooke's law!!! It only requires that the force $F(x)$ is an odd function of the displacement $x$, not a linear function. That is all you need to show that $\displaystyle \int_0^u F(x)\, dx = \int_0^{-u} F(x)\, dx $ for any $u$. | |
| Sep 7, 2017 at 12:34 | history | edited | Emilio Pisanty | CC BY-SA 3.0 |
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| Sep 7, 2017 at 11:09 | answer | added | DarioP | timeline score: 4 | |
| Sep 7, 2017 at 7:48 | history | edited | PinkFloyd | CC BY-SA 3.0 |
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| Sep 7, 2017 at 7:29 | comment | added | PinkFloyd | @Hamsteriffic, you're right I meant that the derivative is discontinuous... This boils down to the same argument that nature "likes smoothness" (at least in classical mechanics) | |
| Sep 7, 2017 at 1:19 | comment | added | Mick | "But if a student argues that kk could be defined with other units so that the dependence in dd is linear" - if integrals are too advanced and they haven't yet covered Hooke's Law how likely is a question like this? I agree with the others, Hooke's Law F = kx, is worth the few seconds. | |
| Sep 7, 2017 at 0:44 | comment | added | Pedro A | What do you mean by a discontinuity in $d = 0$? The function $|d|$ is continuous there, no problem. It is not differentiable... But continuous it is, so you might have to explain it a little further. Hand-waving suggestion: nature does not like "cusps". | |
| Sep 6, 2017 at 19:18 | answer | added | Yakk | timeline score: 11 | |
| Sep 6, 2017 at 19:14 | history | tweeted | twitter.com/StackPhysics/status/905509413993230337 | ||
| Sep 6, 2017 at 14:13 | vote | accept | PinkFloyd | ||
| Sep 6, 2017 at 13:56 | answer | added | JMac | timeline score: 63 | |
| Sep 6, 2017 at 13:55 | answer | added | Gilbert | timeline score: 24 | |
| Sep 6, 2017 at 13:48 | history | protected | Qmechanic♦ | ||
| Sep 6, 2017 at 13:48 | history | edited | Qmechanic♦ | CC BY-SA 3.0 |
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| Sep 6, 2017 at 13:46 | answer | added | Abhinav Dhawan | timeline score: 9 | |
| Sep 6, 2017 at 13:44 | answer | added | Digiproc | timeline score: 19 | |
| Sep 6, 2017 at 13:15 | history | asked | PinkFloyd | CC BY-SA 3.0 |