The potential energy of a 1 kg particle free | vedclass

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(A) The total mechanical energy $E$ is the sum of kinetic energy $K$ and potential energy $U$: $E = K + U = 2 \ J$.To find the maximum speed,we need t

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$\frac{3}{\sqrt{2}}$

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(A) The total mechanical energy $E$ is the sum of kinetic energy $K$ and potential energy $U$: $E = K + U = 2 \ J$.To find the maximum speed,we need to maximize the kinetic energy $K = E - U$.This occurs when the potential energy $U(x)$ is at its minimum.To find the minimum of $U(x)$,we set the derivative $\frac{dU}{dx} = 0$:$\frac{dU}{dx} = \frac{4x^3}{4} - \frac{2x}{2} = x^3 - x = 0$.$x(x^2 - 1) = 0$,which gives $x = 0, 1, -1$.Evaluating $U(x)$ at these points:$U(0) = 0 \ J$.$U(1) = \frac{1}{4} - \frac{1}{2} = -\frac{1}{4} \ J$.$U(-1) = \frac{1}{4} - \frac{1}{2} = -\frac{1}{4} \ J$.The minimum potential energy is $U_{\min} = -\frac{1}{4} \ J$.Thus,$K_{\max} = E - U_{\min} = 2 - (-0.25) = 2.25 \ J = \frac{9}{4} \ J$.Using $K_{\max} = \frac{1}{2}mv_{\max}^2$ with $m = 1 \ kg$:$\frac{1}{2} 	imes 1 	imes v_{\max}^2 = \frac{9}{4}$.$v_{\max}^2 = \frac{9}{2}$.$v_{\max} = \frac{3}{\sqrt{2}} \ m/s$.

The potential energy of a $1 \ kg$ particle free to move along the $x-$axis is given by: $U(x) = (\frac{x^4}{4} - \frac{x^2}{2}) \ J$. The total mechanical energy of the particle is $2 \ J$. Then,the maximum speed (in $m/s$) is:

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