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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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detailed solution

Correct option is A

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)=(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