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.
To find the work done in moving a particle in this potential from x = 1 m to x = 2 m, we need to calculate the change in potential energy. Given that U(x) = 5 - x + 3x^2 - 2x^3, we can find the potential energy at each point and subtract them.
1. Calculate potential energy at x = 1 m:
U(1) = 5 - 1 + 3(1)^2 - 2(1)^3 = 5 + 3 - 2 = 6 J
2. Calculate potential energy at x = 2 m:
U(2) = 5 - 2 + 3(2)^2 - 2(2)^3 = 5 + 12 - 16 = 1 J
The change in potential energy is given by:
ΔU = U(2) - U(1) = 1 J - 6 J = -5 J
The work done in moving the particle from x = 1 m to x = 2 m is equal to the negative of the change in potential energy, so the work done is +5 J.
Next, let's find the force on a particle at x = 1 m and x = 2 m. Force can be calculated as the negative derivative of the potential energy function with respect to x.
1. Force at x = 1 m:
F(1) = -dU(x)/dx at x = 1 m
Taking the derivative of U(x) = 5 - x + 3x^2 - 2x^3 with respect to x:
dU(x)/dx = -1 + 6x - 6x^2
Substituting x = 1:
F(1) = -1 + 6(1) - 6(1)^2 = -1 + 6 - 6 = -1 N
2. Force at x = 2 m:
F(2) = -dU(x)/dx at x = 2 m
Taking the derivative of U(x) = 5 - x + 3x^2 - 2x^3 with respect to x:
dU(x)/dx = -1 + 6x - 6x^2
Substituting x = 2:
F(2) = -1 + 6(2) - 6(2)^2 = -1 + 12 - 24 = -13 N
Therefore, the force on a particle at x = 1 m is -1 N, and at x = 2 m is -13 N.
To locate the points of stable and unstable equilibrium, we need to find the points where the force is zero, i.e., F(x) = 0.
1. Find stable equilibrium point(s):
Set F(x) = 0 and solve for x:
-1 + 6x - 6x^2 = 0
Rearranging the equation:
6x^2 - 6x + 1 = 0
Using the quadratic formula, we find:
x = (6 ± √[(6)^2 - 4(6)(1)]) / (2(6))
x = (6 ± √[36 - 24]) / 12
x = (6 ± √12) / 12
x = (6 ± 2√3) / 12
Simplifying:
x = (1 ± √3) / 6
So the stable equilibrium points are (1 + √3)/6 and (1 - √3)/6.
2. Find unstable equilibrium point:
To find unstable equilibrium points, we need to find the points where the force changes sign. Between the stable equilibrium points, there is an unstable equilibrium point. This point occurs where the force changes from negative to positive.
Given (1 + √3)/6 and (1 - √3)/6 as the stable equilibrium points, the unstable equilibrium point exists between them.
Therefore, the points of stable equilibrium are (1 + √3)/6 and (1 - √3)/6, and the point of unstable equilibrium is between them.