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A particle with total energy E is moving in a potential energy region U(x). Motion of the particle is restricted to the region when?

(A) U(x) > E.

(B) U(x) < E.

(C) U(x) = 0.

(D) U(x) ≤ E.

I think none of the options are correct. According to me, the correct option should be E < 0. Please help me.

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  • $\begingroup$ Perhaps you explain why you think it should be $E < 0$? $\endgroup$
    – kricheli
    Commented Dec 28, 2022 at 11:08
  • $\begingroup$ Because for an object to be bounded to a region of potential energy. The total energy i.e. the sum of kinetic energy and potential energy must be negative. $\endgroup$
    – Gaurav Aggarwal
    Commented Dec 28, 2022 at 11:14
  • $\begingroup$ "It should be $E<0$ because $E$ should be negative." does not really constitute an explanation. ;) Are you aware that absolute values of potential energy are more or less arbitrary? For example, in a problem with gravitational potential energy $mgh$ (massgravitational acceleration on earthheight) it does not really matter where you set the zero height. $\endgroup$
    – kricheli
    Commented Dec 28, 2022 at 11:21
  • $\begingroup$ Let me try to explain again. You can’t have a bounded system with positive charges because potential energy of the system formed by them will always be positive. As a result total energy will also always be positive. But in case of systems of masses and systems formed by positive and negative charges there is an attractive force between them which keeps them from moving away from each other to an infinite distance where they have no influence on each other. The potential energy of these system is taken to be negative. If we want to know if an object will remain bounded we take PE to be 0 at ∞ $\endgroup$
    – Gaurav Aggarwal
    Commented Dec 28, 2022 at 11:45
  • $\begingroup$ Checking for bound states via $E<0$ in the system of two charges is a consequence of the convention to set $U=0$ at infinity and cannot be applied as a general rule. You could modify the potential to $U' = U + U_0$ with $U_0 = $const. and nothing changes except that you check for bound states via the condition $E<U_0$. $\endgroup$
    – kricheli
    Commented Dec 28, 2022 at 11:56

1 Answer 1

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Why not choose D? Because the total energy $$ E = \frac{1}{2}mv^2+U(x)\geq U(x). $$

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  • $\begingroup$ But this is not the condition for boundation. It’s true for all cases of motion in a region of potential energy. $\endgroup$
    – Gaurav Aggarwal
    Commented Dec 28, 2022 at 11:23
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    $\begingroup$ If you read this as a condition for the positions $x$ of the particle, then yes, this is the condition for boundation. Check out a picture like the one top-left in en.wikipedia.org/wiki/Potential_well. If you assign an energy $E$ to the particle, then to satisfy the above inequality the particle must be in the potential well between $x_1$ and $x_2$. $\endgroup$
    – kricheli
    Commented Dec 28, 2022 at 11:31

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