Problem 64 - Calculus Early Transcendentals: Pearson New International Edition

Question

Find the work done by a force \(\mathbf{F}=3 \mathbf{i}-6 \mathbf{j}+7 \mathbf{k}\) pounds in moving an object from \((2,1,3)\) to \((9,4,6)\), where distance is in feet.

Step-by-Step Solution

Answer

The work done is 24 foot-pounds.

Step 1: Understand the Concept of Work

Work is calculated using the dot product formula: \( W = \mathbf{F} \cdot \mathbf{d} \), where \( \mathbf{F} \) is the force vector and \( \mathbf{d} \) is the displacement vector. This requires us to find the displacement vector first and then the dot product.

Step 2: Calculate the Displacement Vector

The displacement vector \( \mathbf{d} \) is the difference between the final and initial position vectors. So, \( \mathbf{d} = (9-2)\mathbf{i} + (4-1)\mathbf{j} + (6-3)\mathbf{k} = 7\mathbf{i} + 3\mathbf{j} + 3\mathbf{k} \).

Step 3: Calculate the Dot Product

To find the work done, calculate the dot product \( \mathbf{F} \cdot \mathbf{d} \). This is given by: \[ W = (3\mathbf{i} - 6\mathbf{j} + 7\mathbf{k}) \cdot (7\mathbf{i} + 3\mathbf{j} + 3\mathbf{k}) \] \[ = (3)\cdot(7) + (-6)\cdot(3) + (7)\cdot(3) \] \[ = 21 - 18 + 21 \] \[ = 24 \].

Step 4: Interpret the Result

The calculated work done by the force is 24 foot-pounds, which matches the unit of work as force is in pounds and distance is in feet.

Key Concepts

Dot Product

In physics, the dot product is a way to multiply two vectors to get a scalar (a simple number, not another vector). It combines the magnitudes of the two vectors and the cosine of the angle between them. The dot product is expressed as: \[ \mathbf{A} \cdot \mathbf{B} = |\mathbf{A}| |\mathbf{B}| \cos(\theta) \] However, when vectors are expressed in component form (like in the example), the dot product is calculated differently. We multiply corresponding components of two vectors and add them up: - For vectors \( \mathbf{A} = a_1 \mathbf{i} + a_2 \mathbf{j} + a_3 \mathbf{k} \) and \( \mathbf{B} = b_1 \mathbf{i} + b_2 \mathbf{j} + b_3 \mathbf{k} \), the formula becomes: \[ \mathbf{A} \cdot \mathbf{B} = a_1 b_1 + a_2 b_2 + a_3 b_3 \] The beauty of the dot product in this scenario is how it simplifies our calculation of work. We use it to figure out how much of the force is acting in the direction of the movement.

Displacement Vector

The displacement vector indicates how far and in what direction an object has moved from its initial position. It's a difference between the initial and final position vectors. In simpler terms, to calculate a displacement vector \( \mathbf{d} \), you subtract the starting coordinates from the ending coordinates:\[ \mathbf{d} = (x_2 - x_1)\mathbf{i} + (y_2 - y_1)\mathbf{j} + (z_2 - z_1)\mathbf{k} \]For example, if an object moves from position \((x_1, y_1, z_1) = (2, 1, 3)\) to position \((x_2, y_2, z_2) = (9, 4, 6)\), the displacement vector would be:- In the x-direction: \( 9 - 2 = 7 \)- In the y-direction: \( 4 - 1 = 3 \)- In the z-direction: \( 6 - 3 = 3 \)Thus, the displacement vector is \( 7\mathbf{i} + 3\mathbf{j} + 3\mathbf{k} \). This tells us exactly how far the object has shifted in each direction.

Vector Calculation

Vector calculations involve operations like addition, subtraction, and multiplication with scalars or other vectors. These calculations help us solve real-world problems in physics, such as finding the work done by a force. To compute a vector's magnitude or to add vectors:- **Addition/Subtraction**: Add or subtract the corresponding components of the vectors. For instance, if vector \( \mathbf{C} = c_1 \mathbf{i} + c_2 \mathbf{j} + c_3 \mathbf{k} \) and vector \( \mathbf{D} = d_1 \mathbf{i} + d_2 \mathbf{j} + d_3 \mathbf{k} \), their sum is: \[ \mathbf{C} + \mathbf{D} = (c_1 + d_1)\mathbf{i} + (c_2 + d_2)\mathbf{j} + (c_3 + d_3)\mathbf{k} \]- **Magnitude**: Determine the length of a vector \( \mathbf{E} = e_1 \mathbf{i} + e_2 \mathbf{j} + e_3 \mathbf{k} \) using: \[ |\mathbf{E}| = \sqrt{e_1^2 + e_2^2 + e_3^2} \]Understanding vector operations helps not only in physics but in many areas involving spatial reasoning. In work-related problems, like the one in our exercise, using vector calculations simplifies finding the result efficiently.