First slide

Concept of potential energy

Question

A particle which is constrained to move along the x-axis, is subjected to a force in the same direction which varies with the distance x of the particle from the origin as $F (x) = - kx + {ax}^{3}$ . Here k and a are positive constants. For $x \geq 0$ , the functional from of the potential energy $U (x)$ of the particle is

Moderate

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Solution

$F = \frac{- dU}{dx} \Rightarrow dU = - F dx$
$\Rightarrow U = - \int_{0}^{x} (- Kx + {ax}^{3}) dx = \frac{{kx}^{2}}{2} - \frac{{ax}^{4}}{4}$
$∴$ We get U = 0 at x = 0 and x = $\sqrt{2 k / a}$
and also U = negative for $x > \sqrt{2 k / a}$ .
So F = 0 at x = 0
i.e. slope of U – x graph is zero at x = 0.

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