Answer to Question #103316 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:

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

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

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

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