Question

The force acting on a particle in one dimension is F = ax - 2xβ³. The corresponding potential energy V(x), assuming V(0) = 0 is given by ______.

Options

• V(x) = ${\displaystyle \alpha x^2-2\beta x^4}$

• V(x) = ${\displaystyle \frac{1}{2}\alpha x^2+\frac{1}{2}\beta x^4}$

• V(x) = ${\displaystyle \alpha x^2+2\beta x^4}$

• V(x) = ${\displaystyle -\frac{1}{2}\alpha x^2-\frac{1}{2}\beta x^4}$

Answer 1

The force acting on a particle in one dimension is F = ax - 2xβ³. The corresponding potential energy V(x), assuming V(0) = 0 is given by ${\displaystyle \underline{\text{V}\left(x\right)=\frac{1}{2}\alpha x^2+\frac{1}{2}\beta x^4}}$.

Explanation:

${\displaystyle {\int }_0^{\text{v}}\text{dv}=-{\int }_0^{\text{v}}\overline{\text{F}}\cdot \text{dx}}$

v = ${\displaystyle -{\int }_0^{\text{v}}(-\alpha x-2\beta x^3)\text{dx}=\frac{\alpha x^2}{2}+\frac{\beta x^4}{2}}$