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 #54837, Physics Mechanics Kinematics Dynamics

The potential energy (in J) of a system in one dimension is given by: $23\mathrm{U}(\mathrm{x}) = 5 - \mathrm{x} + 3\mathrm{x} - 2\mathrm{x}$

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:

Fig.1

The work done in moving a particle in this potential from $x = 1$ m to $x = 2$ m is given by Eq. (1)

$$A = - \left(U (x = 2) - U (x = 1)\right) = - \left[ \left(5 - 2 + 3 \cdot 2 ^ {2} - 2 \cdot 2 ^ {3}\right) - \left(5 - 1 + 3 \cdot 1 ^ {2} - 2 \cdot 1 ^ {3}\right) \right] = 6 A$$

Fig.2

The force on a particle in this potential

$$F (x) = - \frac {\partial U (x)}{\partial x} = - \frac {\partial}{\partial x} (5 - x + 3 x ^ {2} - 2 x ^ {3}) = - (- 1 + 6 x - 6 x ^ {2}) = 1 - 6 x + 6 x ^ {2}$$

The force at $\mathrm{x} = 1\mathrm{\;m}$ and $\mathrm{x} = 2\mathrm{\;m}$

$$F (1) = 1 - 6 \cdot 1 + 6 \cdot 1 ^ {2} = 1 N$$ $$F (2) = 1 - 6 \cdot 2 + 6 \cdot 2 ^ {2} = 1 3 N$$

The points of stable and unstable equilibrium for this system

$$\frac {\partial U (x)}{\partial x} = 0 \Rightarrow \frac {\partial}{\partial x} (5 - x + 3 x ^ {2} - 2 x ^ {3}) = (- 1 + 6 x - 6 x ^ {2}) = - 1 + 6 x - 6 x ^ {2} = 0$$ $$\begin{array}{l} 6x^{2} - 6x + 1 = 0 \Rightarrow D = 6^{2} - 4 \cdot 6 \cdot 1 = 6 \\ x_{1} = \frac{6 + \sqrt{6}}{12} = 0.5 + \frac{\sqrt{6}}{12} \approx 0.704 \\ x_{2} = \frac{6 - \sqrt{6}}{12} = 0.5 - \frac{\sqrt{6}}{12} \approx 0.296 \\ \end{array}$$ $$\begin{array}{l} \frac{\partial^{2} U(x)}{\partial x^{2}} = 6 - 12x \\ \frac{\partial^{2} U(x_{1})}{\partial x^{2}} = -2.448 \Rightarrow \max \text{ (unstable point } x_{1} = \frac{6 + \sqrt{6}}{12} \approx 0.704) \\ \frac{\partial^{2} U(x_{2})}{\partial x^{2}} = +2.448 \Rightarrow \min \text{ (stable point } x_{2} = \frac{6 - \sqrt{6}}{12} \approx 0.296) \\ \end{array}$$