Question

A particle is moving along the \(x\) -axis subject to the potential energy function \(U(x)=a(1 / x)+b x^{2}+c x-d\), where \(a=7.00 \mathrm{~J} \mathrm{~m}, b=10.0 \mathrm{~J} / \mathrm{m}^{2}\), \(c=6.00 \mathrm{~J} / \mathrm{m},\) and \(d=28.0 \mathrm{~J}\) a) Express the force felt by the particle as a function of \(x\). b) Plot this force and the potential energy function. c) Determine the net force on the particle at the coordinate \(x=2.00 \mathrm{~m}\).

Short answer

In summary, the force felt by the particle as a function of \(x\) is given as \(F(x) = - \left(\frac{7.00}{x^2} + 2(10.0)x + 6.00 \right) \mathrm{N}\). When plotted, the force and potential energy functions give a visual representation of their relationship. At the coordinate \(x=2.00 \mathrm{m}\), the net force on the particle is \(-47.75 \mathrm{N}\), which indicates the force acts in the opposite direction of the positive x-axis.

Step-by-step solution

Calculate the derivative of the potential energy function

The force felt by the particle is the negative derivative of the potential energy function with respect to \(x\). We start by finding the derivative of the given potential energy function: \(U(x) = a(\frac{1}{x}) + bx^2 + cx - d\) The derivatives of each term are: \(\frac{d}{dx}(\frac{a}{x}) = -\frac{a}{x^2}\) \(\frac{d}{dx}(bx^2) = 2bx\) \(\frac{d}{dx}(cx) = c\) \(\frac{d}{dx}(-d) = 0\) Now we will sum these derivatives and add the negative sign to get the force function: \(F(x) = - \left(\frac{a}{x^2} + 2bx + c \right)\)

Insert values for a, b, and c

We are given the values \(a=7.00 \mathrm{~J} \mathrm{~m}, b=10.0 \mathrm{~J}/m^2, c=6.00 \mathrm{~J}/m\) and \(d=28.0 \mathrm{~J}\). Using these values, we can write the force function as: \(F(x) = - \left(\frac{7.00}{x^2} + 2(10.0)x + 6.00 \right) \mathrm{N}\)

Plot the force and potential energy functions

In this step, we would use a graphing software or application to plot the force function \(F(x)\) and potential energy function \(U(x)\). This step can't be performed in this text-based format, but you can use tools like Desmos, GeoGebra, or even Microsoft Excel to create the plots. Remember to label the axes and legends for clarity.

Calculate the net force at \(x=2.00 \mathrm{m}\)

Substitute the given coordinate \(x=2.00 \mathrm{m}\) into the force function \(F(x)\) to determine the net force at this point: \(F(2.00) = - \left(\frac{7.00}{(2.00)^2} + 2(10.0)(2.00) + 6.00 \right) \mathrm{N}\) Solving for the net force: \(F(2.00) = - \left(\frac{7.00}{4.00} + 40.0 + 6.00 \right) \mathrm{N} = - \left(1.75 + 46.0 \right) \mathrm{N} = -47.75 \mathrm{N}\) So, the net force on the particle at the coordinate \(x = 2.00 \mathrm{m}\) is \(-47.75 \mathrm{N}\). This indicates the force acts in the opposite direction of the positive x-axis.

Key concepts

Force in Potential Energy Fields

Understanding the relationship between potential energy and the forces in a field is essential for physics students. Potential energy, denoted as U, is the energy by virtue of an object's position relative to other objects. The force experienced by a particle in a potential energy field is directly related to the spatial variation of that potential energy.

For a one-dimensional case, such as a particle moving along the x-axis, the force F the particle feels can be determined by taking the negative gradient of the potential energy function in terms of x. Mathematically, this is represented by the equation F(x) = -dU/dx. This relationship indicates that the force vector points in the direction of decreasing potential energy, attempting to move the particle to a state of lower potential energy and thus more stability.

Consider a particle within a hypothetical field where the potential energy function is given by U(x) = a(1/x) + bx^2 + cx - d. The negative sign in the force equation signifies a key concept: force acts in the opposite direction to the increase in potential energy. In essence, it's as though the particle 'wants' to move towards lower potential energy areas, which corresponds to the intuitive notion of a 'pull' towards energetically favorable positions.

By understanding the concept of force in potential energy fields, students can predict the motion of particles in various physical systems, from celestial bodies to electrons in an electric field.

Derivative of Potential Function

The derivative of a potential function is a fundamental tool in physics used to derive the force experienced by a particle. To better understand this, let's delve into how to find this derivative. The potential energy function in our example is a polynomial of the position x, which makes its derivative straightforward to calculate using basic differentiation rules.

For each term in the potential function such as a/x, bx^2, and cx, we take the derivative with respect to x individually and then combine them. The derivative of a/x is -a/x^2, showing that as the particle moves away from the origin, the force associated with this term decreases. For bx^2, the derivative is 2bx, which implies that the force increases with distance. The derivative of cx is simply the constant c, indicating a constant force contribution regardless of position.

The sum of these derivatives, taken with a negative sign, gives us the force function. This process underscores the importance of calculus in analyzing dynamic systems in physics, where the rate of change in one quantity can determine another vital attribute—in this case, how the rate of change of potential energy with respect to position determines the force on a particle.

Net Force Calculation

Calculating net force is an essential step in resolving the dynamics of a particle in a field. It involves substituting a specific position into our force function to determine the force acting on the particle at that point. In our previous discussion, we derived the force function as F(x) = - (a/x^2 + 2bx + c). To find the net force acting on the particle at any given point, say x=2.00 m, we simply plug this value into the force function.

With the provided values for a, b, c, and d, one can calculate the net force as F(2.00) = - ((7.00/(2.00)^2) + (2*10.0*2.00) + 6.00) Newtons. The resulting calculation yields the net force acting on the particle at this specific location on the x-axis. Understanding net force is paramount as it determines the acceleration of the particle via Newton's second law (F=ma), thereby dictating the particle's subsequent motion.

Net force calculations not only provide snapshots of force at particular positions but can also be employed to map out the complete force profile over a range of positions. This gives deeper insight into the system's dynamics and offers predictive power over the particle’s motion in the potential field.