Timeline for Paradox in simple homework problem
Current License: CC BY-SA 3.0
8 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Jun 13, 2017 at 12:29 | vote | accept | PhyEnthusiast | ||
| May 26, 2017 at 5:08 | comment | added | John Rennie | @PhyEnthusiast: that's the sort of thing we should take to the chat room | |
| May 26, 2017 at 5:07 | comment | added | PhyEnthusiast | ok, got the answer. But, I have another question now. How to calculate that the maximum extension will be $\frac{2mg}{k}$? | |
| May 26, 2017 at 4:58 | history | edited | John Rennie | CC BY-SA 3.0 |
Extend answer
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| May 26, 2017 at 4:46 | comment | added | John Rennie | @PhyEnthusiast: if you do that the mass/spring system oscillates around an average extension of $\Delta mg/k$. When the mass has reached a displacement of $\Delta x$ it has kinetic energy as well as potential energy. The kinetic energy is $mg\Delta x$ minus the work done by the spring $\tfrac{1}{2}k\Delta x^2$. | |
| May 26, 2017 at 4:45 | history | edited | John Rennie | CC BY-SA 3.0 |
Extend answer
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| May 26, 2017 at 4:42 | comment | added | PhyEnthusiast | ok, but what if I hook the block onto the spring and leave the system to reach equilibrium, I do no work on the system. And then what I did is justified [I hope]. | |
| May 26, 2017 at 4:36 | history | answered | John Rennie | CC BY-SA 3.0 |