Answer to Question #102691 in Physics for sakshi

Question

The potential energy (in J) of a system in one dimension is given by: U(x) = 12 - 2x³ + 7x² - 4x. Which are the points of stable and unstable equilibrium for this potential function? Determine the work done in moving a particle in this potential from x = 1 m to x = 2 m

Answer 1

 The work done in moving a particle in this potential from x = 1 m to x = 2 m:

$$W=U(2)-U(1)$$

$$W=(12 - 2(2)^3 + 7(2)^2 - 4(2))-(12 - 2(1)^3 + 7(1)^2 - 4(1))=3\ J$$

For the equilibrium:

$$\frac{dU}{dx}=0=14x-6x^2-4$$

The  points of equilibrium:

$$x_1=\frac{1}{3}, x_2=2$$

$$\frac{d^2U}{dx^2}(x_1)=14-12\frac{1}{3}=10>0$$

$$\frac{d^2U}{dx^2}(x_2)=14-12(2)=-10<0$$

x₁ is the point of stable equilibrium, x₂ is the point of unstable equilibrium.