Problem 50 - Contemporary Precalculus

Question

Find the work done by a constant force \(\boldsymbol{F}\) as the point of application of \(\boldsymbol{F}\) moves along the vector \(\overrightarrow{P Q}\). $$\mathbf{F}=\mathbf{i}-2 \mathbf{j}, P=(0,0), Q=(-5,2)$$

Step-by-Step Solution

Answer

Answer: -9 J

Step 1: Find the displacement vector \(\overrightarrow{PQ}\)

First, we have to find the vector \(\overrightarrow{PQ}\) which represents the displacement from point \(P\) to point \(Q\). Given the coordinates of points \(P(0,0)\) and \(Q(-5,2)\), we can deduce the vector \(\overrightarrow{PQ}\): $$\overrightarrow{PQ} = Q - P = (-5-0)\mathbf{i} + (2-0)\mathbf{j} = -5\mathbf{i} + 2\mathbf{j}$$

Step 2: Calculate the dot product of force \(\boldsymbol{F}\) and displacement vector \(\overrightarrow{PQ}\)

Now we have to calculate the dot product between force vector \(\boldsymbol{F} = \mathbf{i} - 2\mathbf{j}\) and the displacement vector \(\overrightarrow{PQ} = -5\mathbf{i} + 2\mathbf{j}\). The dot product is defined as: $$\boldsymbol{F} \cdot \overrightarrow{PQ} = (\mathbf{i} - 2\mathbf{j}) \cdot (-5\mathbf{i} + 2\mathbf{j})$$ Using the properties of dot product, the expression becomes: $$\boldsymbol{F} \cdot \overrightarrow{PQ} = (\mathbf{i}\cdot -5\mathbf{i}) + (\mathbf{i} \cdot 2\mathbf{j}) + (-2\mathbf{j} \cdot -5\mathbf{i}) + (-2\mathbf{j} \cdot 2\mathbf{j})$$ Now, calculate: $$\mathbf{i} \cdot \mathbf{i} = 1, \quad \mathbf{j} \cdot \mathbf{j} = 1, \quad \mathbf{i} \cdot \mathbf{j} = \mathbf{j} \cdot \mathbf{i} = 0$$ The dot product is now: $$\boldsymbol{F} \cdot \overrightarrow{PQ} = (-5) + 0 + 0 + (-4) = -9$$ The work done by the force \(\boldsymbol{F}\) as the point of application of \(\boldsymbol{F}\) moves along the vector \(\overrightarrow{PQ}\) is -9 J (Joules).

Key Concepts

Dot Product

The dot product, also known as the scalar product, is a fundamental operation in vector mathematics. It involves two vectors and results in a scalar, which is a single number. This operation is crucial in calculating physical quantities like work done by a force, which is why it appears in physics problems.

The dot product for two vectors \( \mathbf{A} = a_1\mathbf{i} + a_2\mathbf{j} \) and \( \mathbf{B} = b_1\mathbf{i} + b_2\mathbf{j} \) in two-dimensional space is given by the formula:
\[ \mathbf{A} \cdot \mathbf{B} = a_1b_1 + a_2b_2 \]

This operation highlights the component-wise multiplication of the vectors followed by their sum.

• The dot product is \( 0 \) if the vectors are perpendicular, indicating no work is done when the force is perpendicular to the displacement.
• If the dot product is positive, the vectors point in roughly the same direction. A negative results in opposite alignment.

Using the properties of the dot product makes it easier to analyze vector components in detailed calculations of work and energy.

Displacement Vector

A displacement vector provides a direction and distance from one point to another in space. It's a key concept in physics, representing how far and in which direction an object has moved. In our given problem, we calculated the displacement vector from point \(P\) to point \(Q\).

To find this vector, we use the coordinates of the two points. If \(P\) has coordinates \((x_1, y_1)\) and \(Q\) has \((x_2, y_2)\), the displacement vector \(\overrightarrow{PQ}\) is:
\[ \overrightarrow{PQ} = (x_2 - x_1)\mathbf{i} + (y_2 - y_1)\mathbf{j} \]

This approach is the foundation of many physics problems, simplifying the process of identifying movement along a path.

• The vector points from the start to the end position, capturing both magnitude and direction.
• Displacement vectors help in describing the change of position effectively, crucial for dynamic calculations.

Understanding displacement aids in solving various problems related to motion and force interactions.

Vector Operations

Vector operations are essential tools in mathematics, physics, and engineering, allowing for manipulation and understanding of vectors in multiple dimensions. These operations include addition, subtraction, dot products, and cross products.

Let's dive into some key operations that you will encounter frequently:

**Addition and Subtraction of Vectors**
Adding or subtracting vectors involves head-to-tail or parallelogram methods. Mathematically, for two vectors \( \mathbf{A} = a_1 \mathbf{i} + a_2 \mathbf{j} \) and \( \mathbf{B} = b_1 \mathbf{i} + b_2 \mathbf{j} \):
\[ \mathbf{A} + \mathbf{B} = (a_1 + b_1) \mathbf{i} + (a_2 + b_2) \mathbf{j} \]
\[ \mathbf{A} - \mathbf{B} = (a_1 - b_1) \mathbf{i} + (a_2 - b_2) \mathbf{j} \]

**Dot Product and Applications**
The dot product, explored previously, provides a measure of vectors' alignment, affecting calculations in work and projections.

The simplicity of operations such as vector addition and the intricacy of products like the dot product make vector operations versatile.

• These operations allow transformation of physical concepts into mathematical expressions.
• By mastering these, you can transition easily between graphical and analytical approaches.

Grasping these operations enhances problem-solving skills in various fields, reflecting both spatial and numerical reasoning.