Question

A force F(x)=(3.0/x)N acts on a particle as it moves along the positive x-axis.

(a) How much work does the force do on the particle as it moves from x=2.0m to x=5.0m?
(b) Picking a convenient reference point of the potential energy to be zero at x=[infinity], find the potential energy for this force.

Answer 1

The work done by the force as the particle moves from x=2.0m to x=5.0m is - (3.0 ln(5.0) - 3.0 ln(2.0)).

Step-by-step explanation:

(a) To find the work done by the force on the particle as it moves from x = 2.0 m to x = 5.0 m, we need to calculate the integral of the force function, F(x), over the given interval. The work done is equal to the negative of the change in potential energy. Work = - ∫ F(x) dx. Substituting the given force function, we have:

Work = - ∫ (3.0/x) dx = - 3.0 ln(x) + C

evaluating the integral from x = 2.0 m to x = 5.0 m:

Work = - (3.0 ln(5.0) - 3.0 ln(2.0))