Problem 34 - Algebra and Trigonometry with Analytic Geometry

Question

A constant force of magnitude 10 has the same direction as \(-\mathbf{i}\). Find the work done if its point of application moves from \(P(0,1)\) to \(Q(1,0)\).

Step-by-Step Solution

Answer

The work done is -10 units.

Step 1: Understand the problem

We need to calculate the work done by a constant force as it moves a point from position \(P(0, 1)\) to position \(Q(1, 0)\). The force has a magnitude of 10 and is directed along \(-\mathbf{i}\).

Step 2: Recall the Work formula

Work done by a force is calculated using the formula \(W = \mathbf{F} \cdot \mathbf{d}\), where \(\mathbf{F}\) is the force vector and \(\mathbf{d}\) is the displacement vector.

Step 3: Determine the displacement vector

The displacement \(\mathbf{d}\) is found by subtracting the coordinates of \(P\) from \(Q\). Thus, \(\mathbf{d} = (1-0)\mathbf{i} + (0-1)\mathbf{j} = \mathbf{i} - \mathbf{j}\).

Step 4: Express the force vector

The force has a magnitude of 10 and is directed along \(-\mathbf{i}\). Therefore, the force vector \(\mathbf{F} = -10\mathbf{i}\).

Step 5: Calculate the dot product

The work done is the dot product of the force and displacement vectors: \(W = \mathbf{F} \cdot \mathbf{d} = (-10\mathbf{i}) \cdot (\mathbf{i} - \mathbf{j})\). This simplifies to \((-10) \cdot 1 + 0 = -10\).

Step 6: Conclude the work done

The work done by the force over the given displacement is \(-10\) units of work, where the negative sign indicates the work is done against the direction of force.

Key Concepts

Understanding Constant Force

In physics, a force is considered constant if it maintains the same magnitude and direction throughout its action. For this particular exercise, we have a force with a magnitude of 10 units. It is crucial to note that this force is aligned in the direction of \(-\mathbf{i}\), meaning it points to the negative x-direction.

A constant force like this doesn't change; hence the calculation of work done becomes straightforward. We do not have to factor in any changes in the force as the object moves. This constancy allows us to use simple vector operations like the dot product to find work done over a specific path.

• Magnitude: 10 units
• Direction: \(-\mathbf{i}\)
• Path: Straight, unchanging influence

Understanding Displacement Vector

The displacement vector represents the change in position of a point from one location to another. Here, we need to determine the path traveled from point \(P(0, 1)\) to point \(Q(1, 0)\).

We find the displacement vector \(\mathbf{d}\) by subtracting the coordinates of the starting point from those of the endpoint:\[\mathbf{d} = (1-0)\mathbf{i} + (0-1)\mathbf{j} = \mathbf{i} - \mathbf{j}\]This vector tells us that the motion includes a move 1 unit to the right (positive x-direction) and 1 unit down (negative y-direction). In summary,

• Start: \(P(0,1)\)
• End: \(Q(1,0)\)
• Displacement: \mathbf{i} - \mathbf{j}\

Explaining the Dot Product

The dot product is a mathematical operation that describes how aligned two vectors are with one another. In this context, we use it to calculate the work done by the force vector on the displacement vector.

The formula for work involves the dot product of the force vector \(\mathbf{F}\) and the displacement vector \(\mathbf{d}\):\[W = \mathbf{F} \cdot \mathbf{d}\]To compute this, multiply the components of the vectors and then sum them:\[W = (-10\mathbf{i}) \cdot (\mathbf{i} - \mathbf{j}) = (-10)(1) + 0 \cdot (-1)\]The result simplifies to a \-10\, indicating that work has been done against the direction of motion, causing the negative sign. The calculation is:

• Component product: \-10 * 1 = \-10\
• Net Work Done: \-10\

Understanding Vector Analysis in Physics

Vector analysis is a fundamental tool in physics to break down and manage physical events involving direction and magnitude. It allows us to calculate work, forces, and movements precisely.

In this exercise, vectors help express a constant force and displacement so we can analyze their interaction effectively. Vectors ensure we understand how these two quantities relate:

• Force Vector: With magnitude and direction, it represents the push or pull.
• Displacement Vector: Shows the path and change in position.
• Dot Product: Connects force and displacement to evaluate work done.

By using vector analysis, physics problems become streamlined, facilitating accurate calculations and a deeper understanding of interactions in motion and force.