The potential energy of a $$1$$ kg particle free to move along the $$x-axis$$ is given by $$V(x)$$ $$=$$ $$x44$$−$$x22$$ Joules. The total Mechanical Energy of the particle $$2J$$. Then the maximum speed is,

Question

The potential energy of a $$1$$ kg particle free to move along the $$x-axis$$ is given by  $$V(x)$$  $$=$$    $$x44$$−$$x22$$  Joules.   The total Mechanical Energy of the particle $$2J$$. Then the maximum speed is,

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total energy $$TE$$$$=2$$ Jfor maximum speed⇒$$KE$$ is maximum, ⇒$$PE$$ is $$minimumPE=V(x)=x44-x22$$  differentiate with respect to x , to find minimum $$PEdVdx=4x34-2x2dVdx=x3-xfor$$ maximum or miinimum $$dVdx=0x3-x=0$$ ⇒$$x2=1$$ ⇒$$x=$$±$$1for$$ minimum value of $$PE$$ consider $$x=-1V(x)=x44-x22substitute$$  $$x=-1V(x)=-14=$$$$PE$$ $$minimumTE=$$$$KE$$$$+PEKEmax=$$$$TE$$$$-PEminKEmax=2--14KEmax=9412mv2=94v=32m/s$$

The potential energy of a 1 kg particle free to move along the x-axis is given by $V (x) = (\frac{x^{4}}{4} - \frac{x 2}{2})$ Joules. The total Mechanical Energy of the particle 2J. Then the maximum speed is,

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The potential energy of a 1 kg particle free to move along the x-axis is given by V(x) = x44−x22 Joules. The total Mechanical Energy of the particle 2J. Then the maximum speed is,

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The potential energy of a 1 kg particle free to move along the x-axis is given by $V (x) = (\frac{x^{4}}{4} - \frac{x 2}{2})$ Joules. The total Mechanical Energy of the particle 2J. Then the maximum speed is,