Problem 37 - Thomas Calculus

Question

Work by a constant force Show that the work done by a constant force field \(\mathbf{F}=a \mathbf{i}+b \mathbf{j}+c \mathbf{k}\) in moving a particle along any path from \(A\) to \(B\) is \(W=\mathbf{F} \cdot \overrightarrow{A B} .\)

Step-by-Step Solution

Answer

The work done is \( W = \mathbf{F} \cdot \overrightarrow{AB} \), calculated using the dot product.

Step 1: Understanding the Problem

We need to prove that the work done by a constant force \( \mathbf{F} = a \mathbf{i} + b \mathbf{j} + c \mathbf{k} \) in moving a particle from a point \( A \) to a point \( B \) can be represented as the dot product of \( \mathbf{F} \) with the vector \( \overrightarrow{AB} \). The work done, \( W \), is given by \( W = \mathbf{F} \cdot \overrightarrow{AB} \).

Step 2: Defining the Displacement Vector

The displacement vector \( \overrightarrow{AB} \) is represented by the position coordinates of points \( A(x_1, y_1, z_1) \) and \( B(x_2, y_2, z_2) \). Therefore, \( \overrightarrow{AB} = (x_2-x_1)\mathbf{i} + (y_2-y_1)\mathbf{j} + (z_2-z_1)\mathbf{k} \).

Step 3: Using the Dot Product Formula

The dot product of two vectors \( \mathbf{F} = a\mathbf{i} + b\mathbf{j} + c\mathbf{k} \) and \( \overrightarrow{AB} = (x_2-x_1)\mathbf{i} + (y_2-y_1)\mathbf{j} + (z_2-z_1)\mathbf{k} \) is calculated as follows:\[ \mathbf{F} \cdot \overrightarrow{AB} = a(x_2 - x_1) + b(y_2 - y_1) + c(z_2 - z_1) \].

Step 4: Expression for Work Done by a Constant Force

The work done by the force \( \mathbf{F} \) is given by the integral of force over the path of motion. However, since \( \mathbf{F} \) is constant, the work done is simply the dot product of \( \mathbf{F} \) with the displacement vector, which simplifies to:\[ W = \mathbf{F} \cdot \overrightarrow{AB} = a(x_2 - x_1) + b(y_2 - y_1) + c(z_2 - z_1) \].

Step 5: Verification of the Result

The result \( W = a(x_2 - x_1) + b(y_2 - y_1) + c(z_2 - z_1) \) aligns with the expression derived from the dot product, confirming that moving the particle along any path between points \( A \) and \( B \) results in the same value for the work done, provided \( \mathbf{F} \) remains constant.

Key Concepts

Dot Product

One of the key concepts in physics, especially when discussing work done by a force, is the dot product. The dot product, also known as the scalar product, is an operation that takes two vectors and returns a scalar (a single number). This operation is denoted by a dot (·) between the vectors, such as \( \mathbf{F} \cdot \overrightarrow{AB} \). When calculating work done, the dot product is crucial because it combines the force vector and the displacement vector efficiently.

The formula for the dot product of two vectors \( \mathbf{F} = a\mathbf{i} + b\mathbf{j} + c\mathbf{k} \) and \( \overrightarrow{AB} = (x_2-x_1)\mathbf{i} + (y_2-y_1)\mathbf{j} + (z_2-z_1)\mathbf{k} \) is:

• Calculate the product of the corresponding components: \( a(x_2 - x_1) + b(y_2 - y_1) + c(z_2 - z_1) \).
• The result is the total work done by the force when the particle moves from \( A \) to \( B \).

This simplicity and power make the dot product an essential tool in physics.

Displacement Vector

The displacement vector \( \overrightarrow{AB} \) is another fundamental aspect when calculating work done. It represents the change in position of a particle from one point to another. In this exercise, the displacement vector is defined using the coordinates of points \( A(x_1, y_1, z_1) \) and \( B(x_2, y_2, z_2) \).

The formula for determining the displacement vector is:

• \( \overrightarrow{AB} = (x_2-x_1)\mathbf{i} + (y_2-y_1)\mathbf{j} + (z_2-z_1)\mathbf{k} \).

This vector signifies not only the direction but also the magnitude (distance) that the particle has traveled. Understanding the displacement vector is crucial as it directly influences how much work is done when the force is applied.

Constant Force Field

A constant force field is one where the force remains the same throughout the movement of an object. In simpler terms, no matter where the particle moves in space, the force vector \( \mathbf{F} \) remains unchanged. This quality is essential for simplifying the calculation of work done because it allows us to use the dot product directly.

In the discussed exercise, the constant force is denoted by \( \mathbf{F} = a \mathbf{i} + b \mathbf{j} + c \mathbf{k} \). This means:

• The force components \( a, b, \) and \( c \) stay constant in the \( x, y, \) and \( z \) directions respectively.

When the force remains constant, we don't need to consider how it might change along the path from \( A \) to \( B \), making the calculation straightforward. In real-world scenarios, this might be like the constant gravitational force on an object near the Earth's surface.