Problem 36 - Thomas Calculus

Question

A revealing experiment By experiment, you find that a force field \(\mathbf{F}\) performs only half as much work in moving an object along path \(C_{1}\) from \(A\) to \(B\) as it does in moving the object along path \(C_{2}\) from \(A\) to \(B\) . What can you conclude about \(\mathbf{F}\) ? Give reasons for your answer.

Step-by-Step Solution

Answer

The force field \( \mathbf{F} \) is path-dependent and not conservative, as work varies between different paths.

Step 1: Understanding Work Done by a Force Field

The work done by a force field \( \mathbf{F} \) on an object moving between two points is given by the line integral \( W = \int_C \mathbf{F} \cdot d\mathbf{r} \), where \( C \) is the path taken. This means work depends on both the force and the path taken from \( A \) to \( B \).

Step 2: Given Conditions in the Problem

The problem states that the work done by the force field \( \mathbf{F} \) along path \( C_1 \) is half of the work done along path \( C_2 \) from \( A \) to \( B \). Let the work done along \( C_1 \) be \( W_1 \) and along \( C_2 \) be \( W_2 \). We have \( W_1 = \frac{1}{2} W_2 \).

Step 3: Drawing Conclusion on the Nature of the Force Field

For conservative force fields, the work done does not depend on the path, only on the initial and final points. Because the work done by \( \mathbf{F} \) is different for paths \( C_1 \) and \( C_2 \), \( \mathbf{F} \) cannot be a conservative force field. It is path-dependent.

Key Concepts

work done

The amount of work done by a force is a fundamental concept in physics. It's defined as the effect of a force moving an object along a path. When applied to a force field like \( \mathbf{F} \), the work done \( W \) is calculated using a line integral:

• The formula for work done is \( W = \int_C \mathbf{F} \cdot d\mathbf{r} \).
• \( \mathbf{F} \) is the vector field representing the force.
• \( d\mathbf{r} \) represents a tiny displacement along path \( C \).

This integral sums all the small amounts of work done along each segment of the path from point \( A \) to point \( B \).
The concept implies that work is not automatically dependent on just the distance between two points, but also on the path taken, as shown by the exercise. Here, differing amounts of work along different paths indicate that the path itself is influencing the work output.

line integrals

Line integrals are an essential tool used to calculate work done in a force field. They integrate a scalar field or vector field over a curve or path, making them perfect for measuring work done as an object moves through a force field.

• In the case of force fields, the vector field \( \mathbf{F} \) is utilized.
• The path \( C \) taken by the object is crucial in determining the result of the integral.

In practice, line integrals involve evaluating expressions like \( \int_C \mathbf{F} \cdot d\mathbf{r} \).
It combines the force applied and the differential distance covered along the path. Evaluating a line integral can show how work is distributed differently along various paths, further demonstrating versatility in analyzing physical scenarios.
In the given exercise, line integrals help illustrate why different paths yield different amounts of work when the force field \( \mathbf{F} \) is not conservative.

conservative force

A conservative force uniquely impacts work because the amount of work done is only related to the initial and final points, not the path taken between them. Characteristics of a conservative force include:

• Path independence: Work done is the same regardless of the trajectory.
• Existence of potential energy: Conservative forces can be associated with potential energies like gravitational or electrostatic forces.
• Reversible paths: If you return to your initial point, net work done is zero.

The given problem points out that \( \mathbf{F} \) does different amounts of work over different paths.
This strongly indicates that \( \mathbf{F} \) is not a conservative force, as it violates path independence.
Conservative forces allow for simpler calculations since only the start and end points matter.
Identifying them provides insights into the energy conservation and mechanical systems.

path-dependent forces

Some forces depend on the path taken, resulting in variable work outputs even between the same two endpoints. Such forces are termed path-dependent forces. These characteristics define path-dependent forces:

• Work is uniquely tied to the specific path taken.
• No associated potential energy is calculated for this force.
• Cannot return to the initial point with zero net work.

Path dependence is prominent in the exercise when the work along path \( C_1 \) differs from path \( C_2 \).
The outcome suggests any energy related to this force cannot be stored or recovered purely through movement.
While not always straightforward to calculate, recognizing path-dependent forces ensures accurate analysis in real-world scenarios, like friction or air resistance. Understanding path-dependent forces equips us with the ability to handle complex situations where path shapes the work outcome.