Problem 35 - Algebra and Trigonometry

Question

Find the work done by the force \(\mathbf{F}\) in moving an object from \(P\) to \(Q .\) $$ \mathbf{F}=4 \mathbf{i}-5 \mathbf{j} ; \quad P(0,0), Q(3,8) $$

Step-by-Step Solution

Answer

The work done is -28 units.

Step 1: Understand the Work Done Formula

The work done by a constant force moving an object from point \( P \) to point \( Q \) 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. The dot (·) represents the dot product of two vectors.

Step 2: Define the Displacement Vector

The displacement vector \( \mathbf{d} \) is calculated by subtracting the position vector of the initial point \( P \) from the position vector of the final point \( Q \). Given \( P(0,0) \) and \( Q(3,8) \), the displacement vector is:\[ \mathbf{d} = (3 - 0) \mathbf{i} + (8 - 0) \mathbf{j} = 3 \mathbf{i} + 8 \mathbf{j} \]

Step 3: Calculate the Dot Product

The dot product \( \mathbf{F} \cdot \mathbf{d} \) is calculated by multiplying the corresponding components of \( \mathbf{F} \) and \( \mathbf{d} \) and then adding those products:\[ \mathbf{F} \cdot \mathbf{d} = (4 \mathbf{i} - 5 \mathbf{j}) \cdot (3 \mathbf{i} + 8 \mathbf{j}) = 4 \times 3 + (-5) \times 8 \]\[ = 12 - 40 = -28 \]

Step 4: Interpret the Result

The result from the dot product calculation gives the work done by the force \( \mathbf{F} \) in moving the object from point \( P \) to point \( Q \). Since the work is negative, it indicates that the force applied did work against the movement.

Key Concepts

Dot Product

The dot product, also known as the scalar product, is a fundamental operation when dealing with vectors. It is used to multiply two vectors, resulting in a scalar quantity. In this exercise, the dot product helps determine the work done by a force. To compute it, multiply the corresponding components of the vectors involved and then sum these products.

Here, the force vector is \(\mathbf{F} = 4 \mathbf{i} - 5 \mathbf{j}\) and the displacement vector is \(\mathbf{d} = 3 \mathbf{i} + 8 \mathbf{j}\). The dot product of these vectors is calculated as follows:

• Multiply the \(\mathbf{i}\) components: \(4 \times 3 = 12\)
• Multiply the \(\mathbf{j}\) components: \(-5 \times 8 = -40\)
• Add these products: \(12 + (-40) = -28\)

This result indicates that the force does \(-28\) units of work. The negative sign suggests that the force acts against the direction of the displacement.

Displacement Vector

The displacement vector signifies the change in position of an object. It is calculated by subtracting the initial position vector from the final position vector. This vector is crucial because it outlines the direction and magnitude of movement from one point to another.

In our exercise, the object moves from initial point \(P(0,0)\) to the final point \(Q(3,8)\). The displacement vector \(\mathbf{d}\) is therefore:

• Subtract the \(x\)-coordinates: \(3 - 0 = 3\)
• Subtract the \(y\)-coordinates: \(8 - 0 = 8\)

Thus, the displacement vector is \(\mathbf{d} = 3 \mathbf{i} + 8 \mathbf{j}\). This vector shows a movement of 3 units in the \(x\)-direction and 8 units in the \(y\)-direction.

Constant Force

A constant force is one that does not change in magnitude or direction during the period of its application. This concept simplifies calculations because the force remains uniform as the object moves from one point to another.

In this problem, the constant force is represented by the vector \(\mathbf{F} = 4 \mathbf{i} - 5 \mathbf{j}\). This means:

• There is a consistent component of force applied parallel to the \(x\)-axis, valued at 4 units.
• There is a steady component of force applied in the negative \(y\)-axis, valued at -5 units.

The uniformity of this force vector makes it easy to apply the work done formula without having to account for changes in direction or magnitude over the pathway from \(P\) to \(Q\).

Vector Components

Vector components are individual parts that combine to form a complete vector. Vectors in physics often have two components in a plane: one along the \(x\)-axis and another along the \(y\)-axis. Understanding these components is essential as they help simplify complex vector operations, like the dot product.

For the force vector \(\mathbf{F} = 4 \mathbf{i} - 5 \mathbf{j}\), the components are:

• 4 \(\mathbf{i}\) is the component along the \(x\)-axis.
• -5 \(\mathbf{j}\) is the component along the \(y\)-axis.

Similarly, the displacement vector \(\mathbf{d} = 3 \mathbf{i} + 8 \mathbf{j}\) consists of:

• 3 \(\mathbf{i}\) as the \(x\)-component.
• 8 \(\mathbf{j}\) as the \(y\)-component.

By breaking down vectors into these components, calculating the dot product becomes a straightforward process of pairing and multiplying respective components.