A particle can move along x axis under the influence of a conservative force. The potential energy of the particle is given by $U = 5 x^{2} - 20 x + 2 (J)$ , where $x$ is coordinate of the particle expressed in $m$ . The particle is released at $x = - 3 m$ . Find the maximum $x$ coordinate of the particle.

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Solution

$U = 5 x^{2} - 20 x + 2$
$F = - \frac{d U}{d x} = 10 x - 20 = 0$ at mean position x=2
Amplitude $= 2 - (- 3) = 5$
Maximum X-coordinate =5+2=7m

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