Q. A particle is placed at the origin and a force $F=kx$ is acting on it (where $k$ is a positive constant). IF $U\left(\right.0\left.\right)=0$ , the graph of $U\left(\right.x\left.\right)$ versus $x$ will be (where $U$ is the potential energy function)

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Solution:

$\text{F} = - \frac{\text{dU}}{\text{dx}} \, \, \, ⇒ \, ∫ \text{dU} = - ∫ \text{Fdx}$
$\text{U}_{2} - \text{U}_{1} = - \left[∫ \text{kxdx}\right] \, \, ⇒ \text{ U} = - \frac{1}{2} \text{kx}^{2}$

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