The potential energy (in J) of a system in one dimension is given by:2 3 U(x) = 5 − x + 3x − 2xWhat is the work done in moving a particle in this potential from x = 1 m to x = 2 m? Whatis the force on a particle in this potentialat x = 1 m and x = 2 m? Locate the points of stableand unstable equilibrium for this system.
To find the work done in moving a particle from x = 1 m to x = 2 m in this potential, we need to calculate the definite integral of the force over the interval [1, 2].
First, let's find the force function by taking the derivative of the potential energy function with respect to x:
U(x) = 5 - x^2 + 3x^2 - 2x^3
F(x) = -dU(x)/dx = -(-2x + 6x^2 - 6x^3)
The force on a particle at x = 1 m and x = 2 m can be calculated by substituting these values into the force function:
F(1) = -(-2+6-6) = 2
F(2) = -(-4+24-48) = -20
Now, to calculate the work done, we will integrate the force function over the interval [1, 2]:
Work = ∫[1, 2] F(x) dx = ∫[1, 2] (-2x + 6x^2 - 6x^3) dx
Integrating term by term, we get:
Work = [-x^2 + 2x^3 - (3/2)x^4] evaluated from x = 1 to x = 2
= (-2 + 16 - 24/2) - (-1 + 2 - (3/2))
= (-2 + 16 - 12) - (-1 + 2 - (3/2))
= 2
Therefore, the work done in moving a particle from x = 1 m to x = 2 m is 2 J.
To find the points of stable and unstable equilibrium, we need to find the points where the force is zero. In other words, we need to solve the equation F(x) = 0.
Set F(x) = 0:
-2x + 6x^2 - 6x^3 = 0
Taking out common factors, we have:
-2x(1 - 3x + 3x^2) = 0
From the equation, we see that one solution is x = 0, which corresponds to the stable equilibrium point.
To find the other points of equilibrium, we solve the quadratic equation 1 - 3x + 3x^2 = 0:
x = (-(-3) ± √((-3)^2 - 4(1)(3)))/(2(3))
= (3 ± √(9 - 12))/6
= (3 ± √(-3))/6
Since the discriminant is negative, there are no real solutions for x in this case. Thus, there are no other equilibrium points for this system.
In summary:
- The work done in moving the particle from x = 1 m to x = 2 m is 2 J.
- The force on the particle at x = 1 m is 2 N, and at x = 2 m is -20 N.
- The point x = 0 is a stable equilibrium point, while there are no other equilibrium points for this system.