Question

A particle of mass $m = 2 kg$ is free to move along $x$ axis under influence of a conservative force. The potential energy function for the particle is given by

$U (x) = a [{(\frac{x}{b})}^{4} - 5 {(\frac{x}{b})}^{2}] J$

Where $a = 1.0 J$ and $b = 1.0 m .$ If the total mechanical energy of the particle is zero,

(a) Plot this potential energy function, identify the regions where particle may be found
(b) Calculate the maximum speed of the particle.

Accepted Answer

Given that

Potential energy $U (x) = a [{(\frac{x}{b})}^{4} - 5 {(\frac{x}{b})}^{2}] J$

Mass of particle $m = 2 kg$

(a) As $a = 1$ and $b = 1$

$∴ U (x) = x^{4} - 5 x^{2} . . . . . (i)$

$U = 0$ at $x = 0$ and $x = \pm \sqrt{5} m$

Slope of $U - x$ curve is given by Differentiating (i) w.r.t $x,$

$\frac{d U}{d x} = 4 x^{3} - 10 x$

For zero slope $\frac{d U}{d x} = 0$ where $x = 0, x = \sqrt{\frac{5}{2}} m, x = - \sqrt{\frac{5}{2}} m$

$x (m)$ $\pm 1$ $\pm \sqrt{\frac{5}{2}}$ $\pm 2$ $\pm 3$

$U (J)$ $- 4$ $- \frac{25}{4}$ $- 4$ $+ 36$

(b) As the total mechanical energy of the particle is zero,

$K + U = E \Rightarrow K + U = 0$

$∴$ The particle will remain between $- \sqrt{5} m$ to $+ \sqrt{5} m .$

The kinetic energy of the particle will be maximum where the potential energy of the particle become minimum.

As $U_{\min} = - \frac{25}{4} J,$ hence $K_{m a x} + (- \frac{25}{4}) = 0 \Rightarrow K_{m a x} = \frac{25}{4} J$

$\Rightarrow K_{m a x} = \frac{1}{2} m v_{m a x}^{2} = \frac{25}{4} J$

$\Rightarrow \frac{1}{2} \times 2 \times v_{m a x}^{2} = \frac{25}{4} J$

$∴ v_{m a x} = \frac{5}{2} m / s$

B M Sharma Solutions for Chapter: Work, Energy and Power, Exercise 49: ILLUSTRATION 8.47

B M Sharma Physics Solutions for Exercise - B M Sharma Solutions for Chapter: Work, Energy and Power, Exercise 49: ILLUSTRATION 8.47

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