Timeline for Is the normal force a conservative force?
Current License: CC BY-SA 3.0
7 events
| when toggle format | what | by | license | comment | |
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| Oct 3, 2014 at 0:57 | comment | added | user4552 | In moving reference systems, the normal is coupled to the force that keeps the object on the surface, so the work will be the same as the one by the force that keeps the object on the surface. This is wrong. The first part is wrong because the values of forces are the same regardless of your frame of reference. The second part is wrong because there doesn't have to be any force that "keeps the object on the surface." Think of a tennis racquet hitting a ball. | |
| Jan 11, 2013 at 14:46 | comment | added | MyUserIsThis | @Alraxite Yes, If you can assign a potential function $V$, then $F=-\nabla V$, and you can do that if and only if $\nabla\times F=0$, so if the curl is 0, then you can assign a potential and the force is conservative. The curl is a trivial calculation for most forces. Particularly, here that force is constant so the curl is 0 and it's conservative. | |
| Jan 11, 2013 at 9:12 | comment | added | Alraxite | Though the reasoning that there is no friction is fine, it's not a satisfactory way to show someone that the work done by the normal force here is conservative. Btw, I do know one way to show if a force is conservative: if we can assign a potential energy function to it. And I'm not sure how do we do that here. | |
| Jan 10, 2013 at 15:57 | comment | added | MyUserIsThis | Btw, if you want a tool to know if a force is conservative, that's the rotational of the force: $\nabla\times F$. Even if you don't understand it yet (I guess you will study it in calculus), if the rotational of a force is 0, then the force is conservative. If it's different from 0, it's not. | |
| Jan 10, 2013 at 15:55 | comment | added | MyUserIsThis | Ok, I see. I would look at this problems not from the point of view of conservative forces, but conservation of energy. Energy is conserved always (universally), which means, the total energy before and after must be the same. If they tell you that there is no friction, then total energy of the system must be conserved, that is, initial potential energy of th block will be divided, part given to kinetic energy of the incline, and part given to kinetic energy of the block. That (friction) was the only way energy could "dissapear", and there's not friction, so energy must be conserved. | |
| Jan 10, 2013 at 15:44 | comment | added | Alraxite | Though I appreciate you writing the answer, I really couldn't understand it! Could you please explain in a less mathematical way that how did you deduce the work done by the block on the incline is conservative here and hence we can use energy conservation to solve this kind of problem? | |
| Jan 10, 2013 at 15:06 | history | answered | MyUserIsThis | CC BY-SA 3.0 |