Work Energy Power

JEE Main level, single correct, medium

135 of 183

Question 135 of 183

The potential energy of a 1 kg1 \mathrm{~kg} particle free move along the x\mathrm{x}-axis is given by
V(x)=(x44−x22)JV(x)=\left(\frac{x^4}{4}-\frac{x^2}{2}\right) J
The total mechanical energy of the particle 2 J2 \mathrm{~J}. Then, the maximum speed (in m/s\mathrm{m} / \mathrm{s} ) is

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Answer: (B) 3/23 / \sqrt{2}

kEmax⁡=ET−Umin⁡\mathrm{k} \mathrm{E}_{\max }=\mathrm{E}_{\mathrm{T}}-\mathrm{U}_{\min }
Umin⁡(±1)=−1/4 J\mathrm{U}_{\min }(\pm 1)=-1 / 4 \mathrm{~J}
KEmax⁡=9/4 J⇒U=32 J\mathrm{KE}_{\max }=9 / 4 \mathrm{~J} \Rightarrow \mathrm{U}=\frac{3}{\sqrt{2}} \mathrm{~J}

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