Problem 28 - University Physics with Modern Physics

Question

In an experiment, one of the forces exerted on a proton is \(\overrightarrow{\boldsymbol{F}}=-\alpha x^{2} \hat{i},\) where \(\alpha=12 \mathrm{N} / \mathrm{m}^{2} .\) (a) How much work does \(\overrightarrow{\boldsymbol{F}}\) do when the proton moves along the straight- line path from the point \((0.10 \mathrm{m}, 0)\) to the point \((0.10 \mathrm{m}, 0.40 \mathrm{m}) ?(\mathrm{b})\) Along the straightline path from the point \((0.10 \mathrm{m}, 0)\) to the point \((0.30 \mathrm{m}, 0) ?\) (c) Along the straight-line path from the point \((0.30 \mathrm{m}, 0)\) to the point \((0.10 \mathrm{m}, 0) ?(\mathrm{d})\) Is the force \(\overrightarrow{\boldsymbol{F}}\) conservative? Explain. If \(\overrightarrow{\boldsymbol{F}}\) 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) -0.104 J; (c) 0.104 J; (d) Not conservative. U(x) = 4x^3.

Step 1: Understand the Force Direction

The force \( \overrightarrow{\boldsymbol{F}} = -\alpha x^2 \hat{i} \) acts only in the \(x\)-direction. Its value depends on the \(x\)-position but not on the \(y\)-position.

Step 2: Solve Part (a) – Path from (0.10 m, 0) to (0.10 m, 0.40 m)

Since the force only has an \(x\)-component, and there is no \(x\)-displacement (since \(x = 0.10\) m remains constant), the work done is zero. Work \(W = \int \overrightarrow{\boldsymbol{F}} \cdot d\overrightarrow{\boldsymbol{s}} = 0\) for a path with no \(x\)-displacement.

Step 3: Solve Part (b) – Path from (0.10 m, 0) to (0.30 m, 0)

The force \( \overrightarrow{\boldsymbol{F}} = -12x^2 \hat{i} \). For work done, \( W = \int_{0.10}^{0.30} -12x^2 \ dx \). Evaluate this integral:\[W = -12 \int_{0.10}^{0.30} x^2 \ dx = -12 \left[ \frac{x^3}{3} \right]_{0.10}^{0.30} = -12 \left( \frac{0.30^3}{3} - \frac{0.10^3}{3} \right)\]Compute accordingly:\[= -12 \left( \frac{0.027}{3} - \frac{0.001}{3} \right) = -12 \left( \frac{0.027 - 0.001}{3} \right) = -12 \times 0.008667 \approx -0.104\,\text{J}\]

Step 4: Solve Part (c) – Path from (0.30 m, 0) to (0.10 m, 0)

Replicating the integral of the opposite path direction:\[W = \int_{0.30}^{0.10} -12x^2 \ dx = -12 \left[ \frac{x^3}{3} \right]_{0.30}^{0.10} = -( -0.104) = +0.104\,\text{J}\]The work done here is the negative of part (b).

Step 5: Determine if Force is Conservative

A force is conservative if the work done is path-independent and only depends on initial and final positions. Analyze the results:1. Work in two different paths between the same endpoints differ,thus \( \overrightarrow{\boldsymbol{F}} \) is not conservative.2. If \( \overrightarrow{\boldsymbol{F}} \) were conservative, potential \( U(x) \) would be derived: \( U(x) = \int -(-12x^2)\,dx = 4x^3 + C \). Set \( C = 0 \) for \( x=0 \).

Key Concepts

Work done by a force

Work is the energy transfer that occurs when a force moves an object over a distance. Imagine a force pushing a box across the floor. If you apply a force and the box moves, work is done. To compute work mathematically, we use the formula: work (\(W\)) equals the integral of the force (\(\overrightarrow{\boldsymbol{F}}\)) applied along the path of displacement (\(d\overrightarrow{\boldsymbol{s}}\)). In simpler terms, this equation can be represented as:\[ W = \int \overrightarrow{\boldsymbol{F}} \cdot d\overrightarrow{\boldsymbol{s}} \]In practical scenarios, if the force has components along specific directions, like in this exercise, we consider only the component of the force in the direction of the displacement for calculating work. If there is no displacement in the direction of the force, as seen in part (a) when the proton moved vertically, the work done is zero since there's no distance covered against the force.

Potential energy in physics

Potential energy is the energy stored in an object due to its position within a force field, like gravity or electromagnetism. It can be thought of as energy waiting to be released. When a conservative force, such as gravity, acts on an object, it can store potential energy, which can be converted into kinetic energy or work later. To understand potential energy in terms of this exercise, think about a ball placed at the top of a hill. Here, the height of the hill is analogous to the position in a force field where energy is stored. Similarly, with the force given in our exercise, a potential energy function, typically denoted as \(U(x)\), would represent the energy stored due to the force \(\overrightarrow{\boldsymbol{F}} = -\alpha x^2 \hat{i}\). The potential energy relates to the conservative forces, calculated by integrating the negative of the force with respect to position, showing how the energy changes as the object changes positions. \[ U(x) = \int -(-12x^2)\,dx = 4x^3 + C \] where \(C\) is the constant of integration, defined by boundary conditions.

Physics integrals

Integrals in physics are vital for calculating quantities like work and potential energy. An integral sums up a series of infinitesimally small values, often representing areas under curves on a graph. In this context, integrals help in understanding how variables like force change with respect to another variable, such as position. The exercise uses definite integrals to compute the work done by evaluating the force over a specific path. For example, in part (b), the integral:\[ W = \int_{0.10}^{0.30} -12x^2 \ dx \]is evaluated to measure the work done when moving from \(x = 0.10\) meters to \(x = 0.30\) meters.By solving this integral, the differences in energy over the given course can be understood, demonstrating how the force affects energy transfer over a path.

Force direction and components

In physics, understanding the direction of forces and their components is crucial. Forces can have different components, each acting in different directions. This means a single force can affect an object differently based on how its components align with the object's movement. In this exercise, force \(\overrightarrow{\boldsymbol{F}} = -\alpha x^2 \hat{i}\) is solely in the \(x\)-direction, indicating that movement in other directions does not influence the work done by this force. In simpler terms, if you push a book across the table in the \(x\)-direction, any force components in the \(y\) or \(z\) directions won’t affect how far the book moves across the table, only the force applied directly along the table impacts movement.Thus, knowing the direction and components of a force allows us to accurately calculate physical outcomes, like work and potential energy, ensuring those calculations reflect real-world interactions.