A particle is moving an $X$-axis has potential energy $U=2-20 x+5 x^2 \mathrm{~J}$ along $X$-axis. The particle is released at $x=-3$. The maximum value of $x$ will be ($x$ is in metre and $U$ is in joules)

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Aicharya · Accepted Answer

Potential energy, $U=2-20 x+5 x^2 J$ When particle gains whole of potential energy in the form of kinetic energy at $x=-3 \mathrm{~m}$, then it will travel maximum distance till whole of its kinetic energy becomes zero. $\begin{array}{ll} \text {i.e., } & K=U=0 \ \Rightarrow & 2-20 x+5 x^2=0 \ & 5 x^2-20 x+2=0 \ \therefore & x=\frac{-(-20) \pm \sqrt{(-20)^2-4 \times 5 \times 2}}{2 \times 5} \ & =\frac{20 \pm \sqrt{360}}{10}=\frac{20 \pm 18.97}{10} \ & x=2 \pm 1.9 \end{array}$ For maximum value of $x$, taking + ve sign $\begin{gathered} x=2+1.9=3.9 \ \therefore \text { Total distance }=|-3|+3.9 \ =6.9 \mathrm{~m} \simeq 7 \mathrm{~m} \end{gathered}$

A particle is moving an \(X\)-axis has potential energy \(U=2-20 x+5 x^2 \mathrm{~J}\) along \(X\)-axis. The…

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