Problem 28 - University Physics with Modern Physics

Question

In an experiment, one of the forces exerted on a proton is \(\overrightarrow{F}\) \(= -a x^2 \hat{\imath}\), where \(\alpha = 12 \mathrm{N/m}^2\). (a) How much work does \(\overrightarrow{F}\) do when the proton moves along the straight-line path from the point (0.10 m, 0) to the point (0.10 m, 0.40 m)? (b) Along the straight-line path from the point (0.10 m, 0) to the point (0.30 m, 0)? (c) Along the straight-line path from the point (0.30 m, 0) to the point (0.10 m, 0)? (d) Is the force \(\overrightarrow{F}\) conservative? Explain. If \(\overrightarrow{F}\) is conservative, what is the potential-energy function for it? Let \(U =\) 0 when \(x =\) 0.

Step-by-Step Solution

Answer

(a) 0 J, (b) -0.10332 J, (c) 0.10332 J, (d) Yes, \( U(x) = 4x^3 \).

Step 1: Understand the Force

The force acting on the proton is given by \( \overrightarrow{F} = -\alpha x^2 \hat{\imath} \). Given \( \alpha = 12 \ \mathrm{N/m}^2 \). The direction of the force is along the \(x\)-axis, and it depends solely on the \(x\)-coordinate.

Step 2: Path from (0.10 m, 0) to (0.10 m, 0.40 m)

Since the force \( \overrightarrow{F} \) only depends on \(x\) and there is no displacement in the \(x\)-direction during the movement from (0.10 m, 0) to (0.10 m, 0.40 m), the work done is \( W = \int F \cdot dx = 0 \).

Step 3: Path from (0.10 m, 0) to (0.30 m, 0)

The displacement is purely in the \(x\)-direction, so calculate the work done by integrating \(W = \int_{0.10}^{0.30} -\alpha x^2 \, dx\). The integral is \(\int_{0.10}^{0.30} -12x^2 \, dx = -\alpha \left[ \frac{x^3}{3} \right]_{0.10}^{0.30}\).Evaluate the definite integral:\(= -12 \left( \frac{0.30^3}{3} - \frac{0.10^3}{3} \right) = -12 \times \left(0.009 - 0.000333\right) = -0.10332 \, \text{J}\).

Step 4: Path from (0.30 m, 0) to (0.10 m, 0)

Calculate the work done from \(0.30\) m back to \(0.10\) m:\(W = \int_{0.30}^{0.10} -12x^2 \, dx = -\alpha \left[ \frac{x^3}{3} \right]_{0.30}^{0.10} = 0.10332 \, \text{J}\).The work truly aligns with expectations, as moving back should require opposite work.

Step 5: Determine if the Force is Conservative

The work done was path-independent: the work for a closed path equals zero \((W_{0.30 \to 0.10} + W_{0.10 \to 0.30} = 0)\). Hence, \(\overrightarrow{F}\) is a conservative force.

Step 6: Identify Potential-Energy Function

Since \( \overrightarrow{F} \) is conservative, the potential energy \( U \) can be found from \( F = -\frac{dU}{dx} = -\alpha x^2 \). Integrate to get the potential-energy function:\( U(x) = \int \alpha x^2 \, dx = \frac{\alpha x^3}{3} + C \).Using \( U = 0 \) at \( x = 0 \), set \( C = 0 \), giving \( U(x) = \frac{12 x^3}{3} = 4x^3 \).

Key Concepts

Conservative Forces

In the world of physics, not all forces behave the same way. Some forces are labeled as \(\text{conservative}\). But what does that mean, exactly? Well, a conservative force is one where the work done in moving a particle between two points does not depend on the path taken. It is only dependent on the starting and ending points. This has an interesting consequence: in a closed path, the net work done by a conservative force is zero. Imagine you move an object from point A to point B and then back to A. If the force is conservative, the total work done over this journey is zero. To determine if a force is conservative, we can look at whether the work done by the force over any closed loop is zero. If so, it assures us the force is conservative. For the force \(\overrightarrow{F} = -\alpha x^2 \hat{\imath}\) given in the exercise, since the paths considered all result in net zero work across closed loops, we can confirm it is indeed a conservative force. This discovery allows us to define something else very easily: a potential energy function.

Potential Energy Function

The concept of potential energy is intimately linked to conservative forces. When a force is conservative, it means we can define a potential energy function \( U(x) \). This function gives us the potential energy at any point, corresponding to that point's position. The relationship between force and potential energy is found through the expression:\[ F = -\frac{dU}{dx} \]This formula tells us that the force is the negative gradient (or derivative) of the potential energy. Thus, knowing the force, we can find the potential energy by integrating. In our exercise, the force \( -\alpha x^2\) leads to a potential energy function through integration:\[ U(x) = \int \alpha x^2 \, dx = \frac{\alpha x^3}{3} + C \]To find the constant \( C \), we use given conditions, such as \( U = 0 \) at \( x = 0 \), which simplifies our function to \( U(x) = 4x^3 \). This relationship is quite helpful as potential energy serves as a measure of stored energy that can convert into other forms, like kinetic energy, during motion.

Integration in Physics

Integration is a powerful mathematical tool used widely in physics. When dealing with forces, especially conservative ones, we're often interested in finding quantities that vary over space or time, such as work done or potential energy. The essence of integration in physics is to sum up small pieces over a continuous range. For instance, if you have a position-dependent force like \( F = -\alpha x^2 \), the work done by this force as an object moves from point \( a \) to point \( b \) can be calculated with a definite integral:\[ W = \int_a^b F \, dx \]Here, integration calculates the total work by summing the infinitesimal contributions (\( F \, dx \)) along the path. The exercise's different paths demonstrated how to carry out these calculations. For the path from 0.10 m to 0.30 m, integration gives \( W = -0.10332 \text{ J} \). It's the detailed integration process that reveals how forces do work under variable conditions.Integration not only helps in assessing work but also in determining potential energy functions from force expressions, solidifying its pivotal role in physics.