Timeline for Springs, elastic potential energy, kinetic energy
Current License: CC BY-SA 3.0
9 events
| when toggle format | what | by | license | comment | |
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| May 26, 2016 at 21:33 | comment | added | Gert | @user3386109: you're trying to make this more complicated than it is. Read the approved answer. It's not complicated. With a Hookean spring there's no energy converted to heat. That's the definition of a Hookean spring. If the OP "fully understands the theory behind a Hookean spring" he wouldn't be asking this basic question. He doesn't want to "go beyond that", he just wants to understand his basic problem. | |
| May 26, 2016 at 21:19 | vote | accept | Vitor Aguiar | ||
| May 26, 2016 at 19:32 | comment | added | user3386109 | @Gert You don't seem to understand the difference between an elastic response and an elastic/inelastic collision. It's clear from the question that the OP fully understands the theory behind a Hookean spring, and wants to go beyond that. | |
| May 26, 2016 at 19:28 | answer | added | M. Enns | timeline score: 4 | |
| May 26, 2016 at 19:20 | comment | added | Gert | A massless, Hookean spring will convert the falling ball's kinetic energy to spring potential energy, until the ball momentarily stops. The restoring force of the spring will then start accelerating the ball upwards, until the spring's potential energy has been fully converted back to ball kinetic energy. You need to look up potential energy of a Hookean spring. | |
| May 26, 2016 at 19:15 | comment | added | Gert | @user3386109: that is not useful and quite confusing to the OP. Even a stiff spring will compress somewhat. That the cause of its elastic response. We're also assuming a Hookean spring: no heat is involved then. | |
| May 26, 2016 at 18:11 | comment | added | user3386109 | I think you have to decide whether it's a ball or a block. If you drop a ball onto a stiff spring, the ball could bounce off without compressing the spring. (Technically that would be an elastic collision.) If you drop a block onto an easily compressed spring, then you're expecting an inelastic collision, i.e. the block stays in contact with the top of the spring. In that case, some of the energy will be converted to heat. So depending on your parameters, the heat loss may or may not be significant, and the kinetic energy of the spring may or may not be significant. | |
| May 26, 2016 at 18:07 | comment | added | garyp | We usually take the spring to be massless, so it has no kinetic energy. I think you need to work on clarifying what you are trying to say. | |
| May 26, 2016 at 17:54 | history | asked | Vitor Aguiar | CC BY-SA 3.0 |