Answer to Question #166944 in Physics for Jay

Question

An object is attracted toward the origin with a force given by F = –(6 Nm-3 )x3 , where F is in Newton's and x is in meters. How much work is done by this force when the object moves from a point 1.5 m away from the origin to a point 4.0 m away from the origin? Assuming this to be a conservative force, what is the change in Potential Energy of the object?

Answer 1

By definition, the work done between two points on x-axis is:

$$A = \int_{x_1}^{x_2}F(x)dx$$

where $x_1 = 1.5, x_2 = 4$, and $F(x) = -6x^3$. Thus, obtain:

$$A = \int_{1.5}^{4}-6x^3dx = -\dfrac{6x^4}{4}\Big|_{1.5}^{4} = \dfrac{6\cdot 1.5^4}{4} - \dfrac{6\cdot 4^4}{4} = -376.40625 \space J$$

The work done by a conservative force is equal to the negative of change in potential energy during that process. Thus, the change in Potential Energy is:

$$\Delta PE = -A = 376.40625 \space J$$

Answer. Work: $-376.40625 \space J$, change in Potential Energy: $376.40625 \space J$.