PhysicsMedium70×since 2002Q8569The potential energy of a 111 kgkgkg particle free to move along the xxx-axis is given by V(x)=(x44−x22)JV\left( x \right) = \left( {{{{x^4}} \over 4} - {{{x^2}} \over 2}} \right)JV(x)=(4x4−2x2)J. The total mechanical energy of the particle is 2J.2J.2J. Then, the maximum speed (in m/sm/sm/s) isA32{3 \over {\sqrt 2 }}23B2{\sqrt 2 }2C12{1 \over {\sqrt 2 }}21D222Check answerSkip