Problem 38 - Precalculus Mathematics for Calculus

Question

Find the work done by the force \(\mathbf{F}\) in moving an object from \(P\) to \(Q\). $$\mathbf{F}=-4 \mathbf{i}+20 \mathbf{j} ; \quad P(0,10), Q(5,25)$$

Step-by-Step Solution

Answer

The work done is 280 units.

Step 1: Understand the Formula for Work Done

The work done by a force in moving an object from one point to another 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 between the initial and final positions of the object. Here, the dot represents the dot product of two vectors.

Step 2: Calculate the Displacement Vector

The displacement vector \( \mathbf{d} \) is found by subtracting the initial position vector \( P(0,10) \) from the final position vector \( Q(5,25) \):\[ \mathbf{d} = (5 - 0) \mathbf{i} + (25 - 10) \mathbf{j} = 5 \mathbf{i} + 15 \mathbf{j} \]

Step 3: Find the Dot Product

The dot product of two vectors \( \mathbf{A} = a_1 \mathbf{i} + a_2 \mathbf{j} \) and \( \mathbf{B} = b_1 \mathbf{i} + b_2 \mathbf{j} \) is given by:\[ \mathbf{A} \cdot \mathbf{B} = a_1b_1 + a_2b_2 \]So, for the vectors \( \mathbf{F} = -4 \mathbf{i} + 20 \mathbf{j} \) and \( \mathbf{d} = 5 \mathbf{i} + 15 \mathbf{j} \), the dot product is:\[ (-4 \times 5) + (20 \times 15) = -20 + 300 = 280 \]

Step 4: Compute the Work Done

Using the result from the dot product, the work done \( W \) by the force \( \mathbf{F} \) in moving the object is:\[ W = 280 \]Thus, the work done is 280 units.

Key Concepts

Force Vector

In physics, a force vector represents the magnitude and direction of a force acting on an object. Typically, it is expressed in terms of the unit vectors \( \mathbf{i} \), \( \mathbf{j} \), and possibly \( \mathbf{k} \) for three-dimensional space. This vector allows us to succinctly describe how and where a force is acting. For instance, consider \( \mathbf{F} = -4 \mathbf{i} + 20 \mathbf{j} \) in this particular problem.

• \( -4 \mathbf{i} \): This component indicates a force acting in the negative x-direction, a push to the left with a magnitude of 4 units.
• \( 20 \mathbf{j} \): This component conveys that there is a force of 20 units towards the positive y-direction, that is upward.

Force vectors are key when calculating work done, as they help determine how much of the force applies in the direction of displacement.

Displacement Vector

A displacement vector tells us how to get from one point to another in space—essentially a directed line segment. This is crucial in determining how much work is done, as it complements the force vector by pointing out where the movement occurs. In our example, we determine the displacement vector by taking the difference between the final and initial position vectors. Starting from point \( P(0,10) \) and ending at \( Q(5,25) \), we form the displacement vector:

• \( \mathbf{d} = (5 - 0) \mathbf{i} + (25 - 10) \mathbf{j} \)
• Simplified, this becomes \( \mathbf{d} = 5 \mathbf{i} + 15 \mathbf{j} \)

This vector shows a movement of 5 units to the right and 15 units up. Knowing both the force and the displacement helps us to compute the actual work involved.

Dot Product

The dot product is a mathematical operation that takes two equal-length sequences of numbers (often coordinate vectors) and returns a single number. It's useful to find out how much of one vector goes in the direction of another. In physics, it's used to calculate work done when both the force vector and the displacement vector are known. For vectors \( \mathbf{A} = a_1 \mathbf{i} + a_2 \mathbf{j} \) and \( \mathbf{B} = b_1 \mathbf{i} + b_2 \mathbf{j} \), the dot product is:
\[ \mathbf{A} \cdot \mathbf{B} = a_1b_1 + a_2b_2 \]
By applying it to the given vectors \( \mathbf{F} = -4 \mathbf{i} + 20 \mathbf{j} \) and \( \mathbf{d} = 5 \mathbf{i} + 15 \mathbf{j} \):

• Calculate the x-components: \(-4 \times 5 = -20\)
• Calculate the y-components: \(20 \times 15 = 300\)
• Add them up: \(-20 + 300 = 280\)

The resulting value, 280, represents the work done in this particular scenario, showing a direct application of the dot product in determining meaningful physical work.