The potential energy of a body of mass 2 kg free to move along x-axis is given by U(x)=(x44−x22) joule. Total mechanical energy of the particle is 2 joule. Then the maximum speed of the body is

F=−dUdx=−(x3−x) At equilibrium position, F=0 ⇒x=±1 ∴Umin=(14−12)=−14J TE=Umin+KEmax KEmax=2.25 12mVmax2=2.25⇒Vmax=1.5ms−1 .

The potential energy of a body of mass 2 kg free to move along x-axis is given by U(x)=(x44−x22) joule. Total mechanical energy of the particle is 2 joule. Then the maximum speed of the body is

F=−dUdx=−(x3−x) At equilibrium position, F=0 ⇒x=±1 ∴Umin=(14−12)=−14J TE=Umin+KEmax KEmax=2.25 12mVmax2=2.25⇒Vmax=1.5ms−1 .

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The potential energy of a body of mass 2 kg free to move along x-axis is given by $U (x) = (\frac{x^{4}}{4} - \frac{x^{2}}{2})$ joule. Total mechanical energy of the particle is 2 joule. Then the maximum speed of the body is

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$1.5 m s^{- 1}$

$2.25 m s^{- 1}$

$\sqrt{2} m s^{- 1}$

$\frac{3}{\sqrt{2}} m s^{- 1}$

answer is A.

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Detailed Solution

$F = - \frac{d U}{d x} = - (x^{3} - x)$

At equilibrium position, F=0

$\Rightarrow x = \pm 1$

$∴ U_{\min} = (\frac{1}{4} - \frac{1}{2}) = - \frac{1}{4} J$

$T E = U_{\min} + K E_{\max}$

$K E_{\max} = 2.25$

$\frac{1}{2} m V_{\max}^{2} = 2.25 \Rightarrow V_{\max} = 1.5 m s^{- 1}$ .

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The potential energy of a body of mass 2 kg free to move along x-axis is given by U(x)=(x44−x22) joule. Total mechanical energy of the particle is 2 joule. Then the maximum speed of the body is

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The potential energy of a body of mass 2 kg free to move along x-axis is given by $U (x) = (\frac{x^{4}}{4} - \frac{x^{2}}{2})$ joule. Total mechanical energy of the particle is 2 joule. Then the maximum speed of the body is