Question

The P.E of a system in one dimension is given by U= 5-x+3x^2-2x^3. What is the work done in moving a particle in this potential from x=1 m to x=2m ? What is the force on the particle in this potential at x=1andx=2 m? Locate the points of stable and unstable equilibrium for this system.

Answer 1

Answer on Question #55877, Physics / Mechanics | Relativity

Task: The P.E of a system in one dimension is given by $U = 5 - x + 3x^{\wedge}2 - 2x^{\wedge}3$ . 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 the particle in this potential at $x = 1$ and $x = 2$ m?

Answer:

The work done in moving the particle in this potential from $x = 1$ m to $x = 2$ m

$$A = U (x = 1) - U (x = 2) = 5 - 1 + 3 - 2 - 5 + 2 - 12 + 16 = 6J$$

the force on the particle in this potential at $x = 1$ and $x = 2$ m:

$$F = - \frac {d U}{d x} = x - 6 x + 6 x ^ {2}$$ $$F (x = 1) = 1 J$$ $$F (x = 2) = 2 - 12 + 24 = 14J$$

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