1D example of the relation between force and potential energy
1DExampleOfTheRelationBetweenForceAndPotentialEnergy
2007-08-12 00:40:15
2007-08-12 00:40:15
Example
relation between force and potential energy
Changes for correction #34 ('1D example of the relation between force and potential energy').
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For a simple one dimensional example of the relationship between force and potential energy, assume that the potential energy of a particle is given by the equation
$$
U(x) = -\frac{A}{x} \left [ 1 + B e^{-x/c} \right]
$$
The realtionship between force and potential energy is
\begin{equation}
\mathbf{F} = -\nabla U
\end{equation}
For our 1D example, where the potential energy is dependent only on the position, $x$
$$ F = -\frac{dU}{dx}$$
Taking the derivative yields
$$ \frac{dU}{dx} = \frac{A}{X^2} \left [ 1 + B e^{-x/c} \right ] - \frac{A}{x} \left [ -\frac{B}{C}e^{-x/c} \right ]$$
Therefore, the force on the particle is governed by the equation
\begin{equation}
F = -\frac{A}{X^2} \left [ 1 + B e^{-x/c} \right ] + \frac{A}{x} \left [ -\frac{B}{C}e^{-x/c} \right ]
\end{equation}
If we further assume that the constant C is so much larger than x, the force will simplify. Rearraning to get
$$
F = -\frac{A}{X^2} \left [ 1 + B e^{-x/c} -\frac{Bx}{C}e^{-x/c} \right ]
$$
Because $x \ll C$, $e^{-x/c} \Rightarrow e^0 = 1$, we get
$$ F = -\frac{A}{x^2} \left [ 1 + B - \frac{Bx}{C} \right ] $$
Further simplifying $\frac{Bx}{C} = 0$ gives the force under this assumption
\begin{equation}
F = -\frac{A}{x^2} \left [ 1 + B \right ]
\end{equation}