Periodic motion -- Potential as a function of a non-linear Force(x)

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SUMMARY

The discussion centers on the potential energy function derived from a non-linear force, specifically defined as f = -dV/dx, where V(x) = -9/x + 18/x^2. The total energy of the system is expressed as e = KE + PE, with KE representing kinetic energy. The user grapples with the implications of negative total mechanical energy (e < 0) and its effect on particle motion, particularly questioning the conditions under which the particle can exist and move, especially as x approaches infinity.

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davon806
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Homework Statement


Please see the attached.I don't know how to do (ai).

potential function is the potential energy defined by f = -dV/dx
e is the total energy of the system where
e = KE + PE
= (dx/dt)^2 /2 + V

Note:m=1 because the particle has a unit mass
If you integrate f,you get V(the PE),which is -9/x + 18/x^2 :
http://www.wolframalpha.com/input/?i=+-+9/x+++18/x^2

Homework Equations

The Attempt at a Solution


I actually sketched the graph as given on the above website,clearly if e is positive then V must be negative,which is shown on the graph when x--->infinity,this is the answer to part (aii).But for a(i) I have no idea,because I haven't met a negative total mechanical energy before.
What I thought is:
KE + V = -e
KE is always positive,leaving V = -e-KE,
This means PE is always negative,but from (aii) I show that V is negative if x--->infinity.In fact,
V = -9/x + 18/x^2 = (18-9x) / x^2.

i.e. V<0 if x>2

If this is the case then in V = -e-Ke we need to have x>2 all the time,which doesn't make sense.Moreover this is the answer to (aii),e>0

Would greatly appreciate if someone can give me an idea what's going on

Thanks
 

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For a given negative energy e<0, where can the particle be? Can it move to infinity? If not, how does its motion have to look like, e. g. if its initial motion is in positive x-direction?
davon806 said:
If this is the case then in V = -e-Ke we need to have x>2 all the time
True. Where is the problem?
 
thx
 

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