Question

The potential energy of a particle experiencing a certain kind of force is given by U(x) = 2x + 8 x measured in joules (J), for positive values of x. What is the minimum potential energy this particle can have?

Answer 1

Answer:

Explanation:

Given

Potential Energy of a particle is given by $U(x)=2x+\frac{8}{x}$

For minimum or maximum Potential Energy differentiate U(x) w.r.t x

$\frac{\mathrm{d} U}{\mathrm{d} x}=2-\frac{8}{x^2}$

putting $\frac{\mathrm{d} }{\mathrm{d} x}=0$

$2-\frac{8}{x^2}=0$

$x^2=\frac{8}{2}$

$x^2=4$

$x=\sqrt{4}=\pm 2$

To know minimum value check $\frac{\mathrm{d^2} U}{\mathrm{d} x^2}$

at x=-2

$\frac{\mathrm{d^2} U}{\mathrm{d} x^2}=\frac{16}{x^3}=-\frac{16}{8}=-2$

so at x=-2 potential energy is minimum $U=-8\ J$