Problem 62 - Calculus Early Transcendentals: Pearson New International Edition

Question

Find the work done by a force \(\mathbf{F}=-5 \mathbf{i}+8 \mathbf{j}\) newtons in moving an object 12 meters north.

Step-by-Step Solution

Answer

The work done is 96 joules.

Step 1: Understanding the Problem

To find the work done by a force, we need to calculate the dot product of the force vector with the displacement vector. The force vector is given as \( \mathbf{F} = -5 \mathbf{i} + 8 \mathbf{j} \) newtons. The object is moved 12 meters north, which means the displacement vector \( \mathbf{s} \) is \( 0 \mathbf{i} + 12 \mathbf{j} \) meters.

Step 2: Calculate the Dot Product

The dot product of 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 \( a_1 b_1 + a_2 b_2 \). For our vectors: \( \mathbf{F} = -5 \mathbf{i} + 8 \mathbf{j} \) and \( \mathbf{s} = 0 \mathbf{i} + 12 \mathbf{j} \), the dot product is \( (-5)(0) + (8)(12) \).

Step 3: Compute the Result

The dot product simplifies to \( 0 + 96 = 96 \). Therefore, the work done by the force in moving the object is 96 joules.

Key Concepts

Dot Product

The dot product is a fundamental operation in vector mathematics frequently used in physics, especially in the context of work and energy. The key idea is that it allows us to measure how much one vector goes in the direction of another vector. This is particularly useful for calculating work done by forces.When calculating the dot product, we are taking two vectors and producing a scalar quantity. For any 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 computed as:\[ a_1 b_1 + a_2 b_2 \]This formula is derived from the cosine of the angle between the two vectors but is simplified when dealing only with coordinates. In the context of work, it determines how much of the force is "effective" in moving the object along the path described by the displacement vector.For the given problem, taking the dot product of the force vector \( \mathbf{F} = -5 \mathbf{i} + 8 \mathbf{j} \) and the displacement vector \( \mathbf{s} = 0 \mathbf{i} + 12 \mathbf{j} \), results in the work done being 96 joules. This operation abstracts away any directions not contributing to the work, focusing only on the effective path.

Force Vector

A force vector describes how force is applied, considering both its direction and magnitude. In physics, forces are represented as vectors because they act in a particular direction and have a certain intensity.For any force \( \mathbf{F} = f_1 \mathbf{i} + f_2 \mathbf{j} \), it combines horizontal and vertical components. These components can include or represent any two overlapping directions, usually denoted by \( \mathbf{i} \) and \( \mathbf{j} \) for the x and y axes, respectively.

• Magnitude: The strength or size of the force.
• Direction: The angle or path it acts along, influencing movement.

In the provided example, the force vector is \( \mathbf{F} = -5 \mathbf{i} + 8 \mathbf{j} \). This signifies the force has a magnitude of 5 units in the negative x-direction and 8 units in the positive y-direction. When calculating work, only the component of force in the direction of displacement matters. Thus, understanding vectors helps to effectively analyze physical situations, leading to accurate calculations of work and energy.

Displacement Vector

The displacement vector provides a straightforward representation of how an object's position changes, incorporating both distance and direction.In mathematical terms, the displacement vector \( \mathbf{s} = s_1 \mathbf{i} + s_2 \mathbf{j} \) describes movement from one point to another in a coordinate system, typically with axes labeled \( \mathbf{i} \) (horizontal) and \( \mathbf{j} \) (vertical).Key features of a displacement vector include:

• Magnitude: The length of the vector, indicating how far the object has moved.
• Direction: The vector's angle or the axis it aligns with, showing where the object goes.

For the exercise, the displacement vector is \( \mathbf{s} = 0 \mathbf{i} + 12 \mathbf{j} \), which indicates a 12-meter movement straight north. This direction opposes the horizontal \( \mathbf{i} \) vector and fully utilizes the vertical \( \mathbf{j} \) component. By focusing on this vector, you can determine only the portion of the force vector that contributes to work, emphasizing the straight-line distance in the given direction.