Calculating Potential Energy from Force F(x) = AX² - Bx

The problem asks us to find the potential energy of a particle given the force acting on it along the X-axis. The force is given as a function of position x, specifically F(x) = AX² - Bx.

In physics, for a conservative force acting along one dimension (like the X-axis), the relationship between the force F(x) and the potential energy U(x) is given by:

$$F(x) = -\frac{dU}{dx}$$

To find the potential energy U(x) from the force F(x), we need to integrate the force with respect to x, and include a negative sign:

$$dU = -F(x) \, dx$$

$$U(x) = \int -F(x) \, dx$$

Substitute the given expression for F(x):

$$U(x) = \int -(AX^2 - Bx) \, dx$$

$$U(x) = \int (-AX^2 + Bx) \, dx$$

Now, we integrate term by term:

$$U(x) = \int -AX^2 \, dx + \int Bx \, dx$$

We can pull the constants A and B out of the integrals:

$$U(x) = -A \int X^2 \, dx + B \int x \, dx$$

Using the power rule for integration (\(\int x^n \, dx = \frac{x^{n+1}}{n+1} + C\), where C is the integration constant):

$$U(x) = -A \left(\frac{X^{2+1}}{2+1}\right) + B \left(\frac{x^{1+1}}{1+1}\right) + C$$

$$U(x) = -A \frac{X^3}{3} + B \frac{x^2}{2} + C$$

This is the general expression for the potential energy, including the integration constant C. The value of C depends on where the zero point of potential energy is defined. Often, the potential energy is defined to be zero at some reference point, such as x=0 or x=\(\infty\). Looking at the provided options, none include an arbitrary constant C. This suggests that the potential energy is likely defined such that U(0) = 0. If U(0) = 0, then substituting x=0 into our expression:

$$U(0) = -A \frac{(0)^3}{3} + B \frac{(0)^2}{2} + C = 0 + 0 + C = C$$

So, if U(0) = 0, then C = 0. In this case, the potential energy function becomes:

$$U(x) = -\frac{Ax^3}{3} + \frac{Bx^2}{2}$$

Now let's compare this form with the given options. The second option is given as \(-\frac{{{x^2}}}{6}\left( {2Ax - 3B} \right)\). Let's expand this expression to see if it matches our result:

$$-\frac{x^2}{6}(2Ax - 3B) = -\frac{x^2}{6}(2Ax) - \frac{x^2}{6}(-3B)$$

$$= -\frac{2Ax^3}{6} + \frac{3Bx^2}{6}$$

$$= -\frac{Ax^3}{3} + \frac{Bx^2}{2}$$

This expanded form exactly matches the potential energy function we derived assuming U(0)=0. Therefore, the potential energy of the particle is given by the second option.

Revision Table: Force and Potential Energy

Additional Information on Potential Energy Concepts

Potential energy is a very useful concept when dealing with conservative forces. It simplifies calculations involving work and energy, especially when forces vary with position.

• The change in potential energy (\(\Delta U\)) of a system is defined as the negative of the work done by the conservative force within the system when the configuration changes. \(\Delta U = -W_c\).
• The concept of potential energy is only meaningful for conservative forces. Friction, for instance, is a non-conservative force, and there is no associated potential energy function for it.
• The total mechanical energy (\(E = K + U\)), which is the sum of kinetic energy (K) and potential energy (U), is conserved if only conservative forces are doing work on the system.
• The specific value of potential energy at a point depends on the choice of the reference point where the potential energy is defined to be zero. However, the change in potential energy (\(\Delta U\)) between two points is independent of this choice.
• In this problem, since the options provide a specific functional form without an additive constant, it is implied that the reference point for potential energy is chosen such that \(U(x)=0\) at \(x=0\).