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GR8677 #34
Problem
 GREPhysics.NET Official Solution Alternate Solutions

Mechanics$\Rightarrow$}Potential Energy

Recall the lovely relation,

$F_x=-\frac{\partial U}{\partial x} \Leftrightarrow \vec{F}=-\nabla U.
$

$U=kx^4\Rightarrow F=-4kx^3$

Alternate Solutions
 casseverhart132019-09-27 02:58:26 Well this is a real life-saver! commercial cleaningReply to this comment TimToolMan2018-04-03 00:15:02 Simple Unit argument.\\\\r\\\\nEnergy is in units of $kg(m^2/ s^2)$\\\\r\\\\nSo we can get the units of k to be $kg/(m^2 s^2)$\\\\r\\\\nForce must be in units of $kg (m/s^2)$\\\\r\\\\nKeeping k\\\\\\\'s units in mind, there are only 2 solutions with these units of force: B & E\\\\r\\\\nNo gravity in this question, so it must be B Reply to this comment
casseverhart13
2019-09-27 02:58:26
Well this is a real life-saver! commercial cleaning
joshuaprice153
2019-08-09 03:07:39
With more intentions to get back to your answer, i was annoyed last week when i didn\'t even got time to read anything over internet. mobile window tinting
TimToolMan
2018-04-03 00:15:02
Simple Unit argument.\\\\r\\\\nEnergy is in units of $kg(m^2/ s^2)$\\\\r\\\\nSo we can get the units of k to be $kg/(m^2 s^2)$\\\\r\\\\nForce must be in units of $kg (m/s^2)$\\\\r\\\\nKeeping k\\\\\\\'s units in mind, there are only 2 solutions with these units of force: B & E\\\\r\\\\nNo gravity in this question, so it must be B
OrrinJelo
2009-08-04 10:37:53
The easiest way is by doing dimensional analysis. (A) is in units of energy, same with (C). (D) is energy times distance. This leaves (B) and (E). (E) is the force associated with gravity, which our potential energy equation looks nothing like the gravitational potential equation. We are left with (B).
 neon372010-11-01 10:42:12 I think thats the hardest way to do this. Easiest is the solution given. You should remember that $\vec{F}= - \nabla{U}$. If you dont then you should think about memorizing that one. It will be mighty helpful.
 pam d2011-09-24 09:53:55 neon37, I second that.
 mpdude82012-04-15 21:03:04 Even though that's the long route for this problem, knowing the units of "k" will be useful in 35 and 36.
jitin1984
2006-10-27 15:15:32
hey

i cant get enough practice problems for quant. i have got schaum seeries but it has only derivations any one has any material plz i want it
Mexicorn
2005-11-08 12:10:14
I think you mean $U=kx^4$ instead of $U=kx^5$ for the potential given.
 yosun2005-11-09 02:27:05 Mexicorn: thanks for the typo-alert; it has been corrected.

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