The PE of a 2 , kg particle,free to move alon | vedclass

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(D) The potential energy is given by $U(x) = \frac{x^3}{3} - \frac{x^2}{2}$.To find the equilibrium points,we set the force $F = -\frac{dU}{dx} = 0$.$

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$\frac{5}{\sqrt{6}}$

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(D) The potential energy is given by $U(x) = \frac{x^3}{3} - \frac{x^2}{2}$.To find the equilibrium points,we set the force $F = -\frac{dU}{dx} = 0$.$\frac{dU}{dx} = x^2 - x = 0 \implies x(x - 1) = 0$,so $x = 0$ or $x = 1$.We check the second derivative $\frac{d^2U}{dx^2} = 2x - 1$.At $x = 0$,$\frac{d^2U}{dx^2} = -1 At $x = 1$,$\frac{d^2U}{dx^2} = 1 > 0$ (stable equilibrium,minimum potential energy).The minimum potential energy is $U_{min} = \frac{1^3}{3} - \frac{1^2}{2} = \frac{1}{3} - \frac{1}{2} = -\frac{1}{6} \, J$.The total mechanical energy $E = K + U = 4 \, J$.Maximum kinetic energy $K_{max}$ occurs when $U$ is minimum: $K_{max} = E - U_{min} = 4 - (-\frac{1}{6}) = \frac{25}{6} \, J$.Using $K_{max} = \frac{1}{2} m v_{max}^2$ with $m = 2 \, kg$:$\frac{1}{2} (2) v_{max}^2 = \frac{25}{6} \implies v_{max}^2 = \frac{25}{6} \implies v_{max} = \frac{5}{\sqrt{6}} \, m \, s^{-1}$.

The $PE$ of a $2 \, kg$ particle,free to move along the $x$-axis,is given by $V(x) = \left( \frac{x^3}{3} - \frac{x^2}{2} \right) \, J$. The total mechanical energy of the particle is $4 \, J$. The maximum speed (in $m \, s^{-1}$) is ...........

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