The potential energy of a $1 \mathrm{~kg}$ particle free move along the $\mathrm{x}$-axis is given by $…

The potential energy of a $1 \mathrm{~kg}$ particle free move along the $\mathrm{x}$-axis is given by $ V(x)=\left(\frac{x^4}{4}-\frac{x^2}{2}\right) J $ The total mechanical energy of the particle $2 \mathrm{~J}$. Then, the maximum speed (in $\mathrm{m} / \mathrm{s}$ ) is
  1. 2
  2. $3 / \sqrt{2}$
  3. $\sqrt{2}$
  4. $1 / \sqrt{2}$

Solution

$\mathrm{k} \mathrm{E}_{\max }=\mathrm{E}_{\mathrm{T}}-\mathrm{U}_{\min }$ $\mathrm{U}_{\min }(\pm 1)=-1 / 4 \mathrm{~J}$ $\mathrm{KE}_{\max }=9 / 4 \mathrm{~J} \Rightarrow \mathrm{U}=\frac{3}{\sqrt{2}} \mathrm{~J}$

Asked in: JEE Main 2006

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