# The potential energy for a certain mass moving in one dimension is given by U(x) = (2.0 J/m^3)x^3 - (15 J/m^2)x^2 + (36 J/m)x - (23 J). Find the location(s) where the force on the mass is zero.

## Expert Answers

The potential energy is given by

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

The force with which this potential energy is associated is

`F = -(dU)/(dx)`

Find the derivative of U with respect to x:

`(dU)/(dx) = 6x^2-30x + 36`

The force is zero when the function above...

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The potential energy is given by

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

The force with which this potential energy is associated is

`F = -(dU)/(dx)`

Find the derivative of U with respect to x:

`(dU)/(dx) = 6x^2-30x + 36`

The force is zero when the function above is 0, that is,

`6x^2 - 30x + 36 = 0`

Divide by 6:

`x^2 - 5x + 6 = 0`

Solve by factoring:

(x-2)(x-3) = 0

x = 2 and x = 3

At x = 2 and x = 3 the force on the mass will be 0.

Approved by eNotes Editorial Team