Problem 30 - Algebra and Trigonometry

Question

29-32 Find the work done by the force \(\mathbf{F}\) in moving an object from \(P\) to \(Q .\) $$\mathbf{F}=400 \mathbf{i}+50 \mathbf{j} ; \quad P(-1,1), Q(200,1)$$

Step-by-Step Solution

Answer

The work done is 80400 units.

Step 1: Understand the Problem Context

The problem asks us to find the work done by a force vector \( \mathbf{F} = 400 \mathbf{i} + 50 \mathbf{j} \) as it moves an object in a straight line path from point \( P(-1, 1) \) to point \( Q(200, 1) \).

Step 2: Recall the Work Formula

The work \( W \) done by a constant force \( \mathbf{F} \) moving an object along a displacement \( \mathbf{d} \) is calculated using the dot product: \[ W = \mathbf{F} \cdot \mathbf{d} \] where \( \mathbf{d} \) is the displacement vector from \( P \) to \( Q \).

Step 3: Calculate the Displacement Vector

Displacement \( \mathbf{d} \) from \( P(-1, 1) \) to \( Q(200, 1) \) can be found by subtracting coordinates:\[ \mathbf{d} = \langle 200 - (-1), 1 - 1 \rangle = \langle 201, 0 \rangle \]

Step 4: Compute the Dot Product

Calculate the dot product \( \mathbf{F} \cdot \mathbf{d} \): \[ \mathbf{F} \cdot \mathbf{d} = (400)(201) + (50)(0) = 80400 + 0 = 80400 \]

Step 5: Interpret the Result

The work done by the force \( \mathbf{F} \) in moving the object from point \( P \) to point \( Q \) is 80400. This value is a scalar quantity, measured in joules if the force was in newtons and distance in meters.

Key Concepts

Dot Product

The dot product is a mathematical operation that combines two vectors and results in a scalar. This means that we have a single number as the outcome, not another vector. Understanding the dot product can help us see how two vectors interact.
To calculate the dot product of two vectors, such as force and displacement, we break each vector into components. If we have two vectors \( \mathbf{A} = a_1 \mathbf{i} + a_2 \mathbf{j} \) and \( \mathbf{B} = b_1 \mathbf{i} + b_2 \mathbf{j} \), then their dot product is calculated as:

• \( \mathbf{A} \cdot \mathbf{B} = a_1b_1 + a_2b_2 \)

In the original example, the dot product of the force vector \( \mathbf{F} = 400 \mathbf{i} + 50 \mathbf{j} \) and the displacement vector \( \mathbf{d} = 201 \mathbf{i} + 0 \mathbf{j} \) illustrates this concept. The dot product results in the calculation \( 400 \times 201 + 50 \times 0 = 80400 \).
The dot product is especially useful in physics to find work done as it combines force and movement into a single meaningful metric.

Displacement Vector

A displacement vector provides direction and distance from one point to another. It tells us how far an object has moved and in which direction.
To calculate a displacement vector between two points, \( P(x_1, y_1) \) and \( Q(x_2, y_2) \), we subtract the initial coordinates from the final coordinates:

• \( \mathbf{d} = \langle x_2 - x_1, y_2 - y_1 \rangle \)

In our problem, the displacement vector from \( P(-1, 1) \) to \( Q(200, 1) \) is given by:

• \( \mathbf{d} = \langle 200 - (-1), 1 - 1 \rangle = \langle 201, 0 \rangle \)

This shows that the object only moves horizontally, 201 units to the right. The magnitude and direction of the displacement vector are crucial, as they directly affect the calculation of work done when combined with the force vector.

Force Vector

A force vector represents the direction and magnitude of a force acting on an object. In physics, forces are described using vectors since they display both size and orientation.
The initial problem provides the force vector \( \mathbf{F} = 400 \mathbf{i} + 50 \mathbf{j} \). Here, "400" acts in the horizontal (\( \mathbf{i} \)) direction, and "50" acts in the vertical (\( \mathbf{j} \)) direction. This means the force applied is stronger horizontally than vertically.
Understanding force vectors is crucial because they indicate how an object should theoretically move. Combined with the displacement vector, we use the force vector in calculations to find the work done. It highlights the relationship between how much force we apply and the motion achieved.

Calculation of Work in Physics

In physics, calculating work helps us understand the energy required to move an object using force. The concept of work ties energy, force, and displacement together beautifully.
Work, denoted as \( W \), is calculated using the formula involving the dot product:

• \( W = \mathbf{F} \cdot \mathbf{d} \)

Where \( \mathbf{F} \) is the force vector, and \( \mathbf{d} \) is the displacement vector. In our example, the calculation resulted in \( W = 80400 \).
This tells us that an amount of energy equal to 80400 (in joules, if measured in standard units like newtons and meters) is used. Calculating work helps us gauge how much effort is required to shift an object from one location to another, crucial for understanding mechanical and energy systems.