Question

The potential energy for a certain mass moving in one dimension is given by U(x)=(2.0J/m3)x3−(15J/m2)x2+(36J/m)x−23JU(x)=(2.0J/m3)x3−(15J/m2)x2+(36J/m)x−23JU(x) = (2.0 {\rm J/m}^{3})x^{3}- (15 {\rm J/m}^{2})x^{2}+ (36 {\rm J/m})x - 23 {\rm J}. Find the location(s) where the force on the mass is zero.

Answer 1

Answer:x=2 and x=3

Explanation:

Given

Potential Energy for a certain mass is

$U(x)=2x^3-15x^2+36x-23$

and we know force is given by

$F=-\frac{\mathrm{d} U}{\mathrm{d} x}$

$F=-(2\times 3x^2-15\times 2x+36)$

For Force to be zero F=0

$\Rightarrow 6x^2-30x+36=0$

$\Rightarrow x^2-5x+6=0$

$\Rightarrow x^2-2x-3x+6=0$

$\Rightarrow (x-2)(x-3)=0$

Therefore at x=2 and x=3 Force on particle is zero.