Question

The potential energy (in J) of a system in one dimension is given by:

2 3 U (x) = 5 − x + 3x − 2x
What is the work done in moving a particle in this potential from x = 1 m to x = 2 m? What
is the force on a particle in this potential at x = 1 m and x = 2 m? Locate the points of stable
and unstable equilibrium for this system.

Answer 1

Answer on Question#54939 – Physics – Mechanics | Kinematics | Dynamics

$$\Delta U = -6 J$$ $$F(1) = 1 N$$ $$F(2) = 13 N$$ $$\left(\frac{1}{2} - \frac{1}{2\sqrt{3}}\right) \text{ m - point of stable equilibrium}$$ $$\left(\frac{1}{2} + \frac{1}{2\sqrt{3}}\right) \text{ m - point of unstable equilibrium}$$

Question

The potential energy (in $J$) of a system in one dimension is given by:

$$U(x) = 5 - x + 3x^2 - 2x^3$$

What is the work done in moving a particle in this potential from $x = 1$ m to $x = 2$ m? What is the force on a particle in this potential at $x = 1$ m and $x = 2$ m? Locate the points of stable and unstable equilibrium for this system.

Solution

Work done: $\Delta U = U(2) - U(1) = (5 - 2 + 3 * 2^2 - 2 * 2^3) - (5 - 1 + 3 * 1^2 - 2 * 1^3) =$

$$(5 - 2 + 12 - 16) - (5 - 1 + 3 - 2) = -1 - 5 = -6 (J)$$

Note: potential energy at the end point is lower than at the start point. Negative difference of energies means that particle get some additional kinetic energy while moving in potential.

Force acting on a particle: $F = -\frac{dU}{dx} = 1 - 6x + 6x^2$

$$F(1) = 1 - 6 * 1 + 6 * 1^2 = 1 - 6 + 6 = 1 (N)$$ $$F(2) = 1 - 6 * 2 + 6 * 2^2 = 1 - 12 + 24 = 13 (N)$$

Points of equilibrium $\equiv F = 0 (N)$: $1 - 6x + 6x^2 = 0 \rightarrow x^2 - x + \frac{1}{6} = 0$

$$D = 1 - \frac{4}{6} = \frac{2}{6} = \frac{1}{3}$$ $$x_1 = \frac{1 - \sqrt{\frac{1}{3}}}{2} = \frac{1}{2} - \frac{1}{2\sqrt{3}}$$ $$x_2 = \frac{1 + \sqrt{\frac{1}{3}}}{2} = \frac{1}{2} + \frac{1}{2\sqrt{3}}$$

In case of minimum of potential at the point we have stable equilibrium, in case of maximum – unstable.

Therefore, we use second derivative test to determine it:

$$\frac {d ^ {2} U}{d x ^ {2}} = \frac {d}{d x} (- 1 + 6 x - 6 x ^ {2}) = 6 - 1 2 x = 6 (1 - 2 x)$$ $$\frac {d ^ {2} U}{d x ^ {2}} \big | _ {x _ {1}} = 6 \left(1 - 2 * \left(\frac {1}{2} - \frac {1}{2 \sqrt {3}}\right)\right) = 6 \left(\frac {1}{2 \sqrt {3}}\right) = \sqrt {3} > 0$$ $$\frac {d ^ {2} U}{d x ^ {2}} \big | _ {x _ {2}} = 6 \left(1 - 2 * \left(\frac {1}{2} + \frac {1}{2 \sqrt {3}}\right)\right) = 6 \left(- \frac {1}{2 \sqrt {3}}\right) = - \sqrt {3} < 0$$

Thus, at $\left(\frac{1}{2} -\frac{1}{2\sqrt{3}}\right)$ ( $m$ ) we have stable equilibrium, and at $\left(\frac{1}{2} +\frac{1}{2\sqrt{3}}\right)$ ( $m$ ) we have unstable equilibrium.

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