contestada

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.

Respuesta :

Answer:x=2 and x=3

Explanation:

Given

Potential Energy for a certain mass is

[tex]U(x)=2x^3-15x^2+36x-23[/tex]

and we know force is given by

[tex]F=-\frac{\mathrm{d} U}{\mathrm{d} x}[/tex]

[tex]F=-(2\times 3x^2-15\times 2x+36)[/tex]

For Force to be zero F=0

[tex]\Rightarrow 6x^2-30x+36=0[/tex]

[tex]\Rightarrow x^2-5x+6=0[/tex]

[tex]\Rightarrow x^2-2x-3x+6=0[/tex]

[tex]\Rightarrow (x-2)(x-3)=0[/tex]

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

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