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JEE PHYSICS MCQ: WORK-ENERGY-POWER

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The potential energy of a particle of mass 3 kg3 {~kg} free to move along the xx-axis is given by U(x)=(3x22−2x)JU(x)=\left(\frac{3 x^2}{2}-2 x\right) {J}. If total mechanical energy of the particle is 5 J5 {~J}, then the maximum speed of the particle is;
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Correct option: B
343 m/s\frac{\sqrt{34}}{3} {~m} / {s}
Option B (343 m/s\frac{\sqrt{34}}{3} {~m} / {s}) is correct because it is calculated by finding the minimum potential energy and then using conservation of energy to find the maximum kinetic energy and thus the maximum speed.

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