Question

Work by a constant force Evaluate a line integral to show that the work done in moving an object from point \(A\) to point \(B\) in the presence of a constant force \(\mathbf{F}=\langle a, b, c\rangle\) is \(\mathbf{F} \cdot \overrightarrow{A B}\).

Short answer

Question: Determine the work done in moving an object from point A to point B in the presence of a constant force 𝐅. Answer: The work done in moving the object from point A to point B in the presence of a constant force 𝐅 is equal to the dot product of the force and the displacement vector, given by: 𝐅 ⋅ 𝑨𝑩 = a(x₂-x₁) + b(y₂-y₁) + c(z₂-z₁).

Step-by-step solution

Parameterize the path from point A to point B

Let \(A = \langle x_1, y_1, z_1 \rangle\) and \(B = \langle x_2, y_2, z_2 \rangle\). We need to parameterize the path from A to B. Consider the straight-line path between points \(A\) and \(B\). To parameterize this path with respect to a single parameter \(t\), observe that for \(t=0\), the path should be at point \(A\), while for \(t=1\), the path should be at point \(B\). This can be expressed by the parameterization: $$\mathbf{r}(t) = \langle x_1 + t(x_2 - x_1), y_1 + t(y_2 - y_1), z_1 + t(z_2 - z_1) \rangle$$

Compute the derivative of the path with respect to the parameter

In order to compute the line integral, we need to find the derivative of the parameterized path with respect to the parameter \(t\). We take the derivative component-wise: $$\frac{d\mathbf{r}(t)}{dt} = \langle x_2 - x_1, y_2 - y_1, z_2 - z_1 \rangle$$

Evaluate the dot product of the force and the derivative of the path

The work done by the constant force \(\mathbf{F}\) along the path is given by the line integral: $$\int_{0}^{1} \mathbf{F} \cdot \frac{d\mathbf{r}(t)}{dt} \, dt$$ To compute this integral, we first evaluate the dot product of \(\mathbf{F}\) and the derivative of the path: $$\mathbf{F} \cdot \frac{d\mathbf{r}(t)}{dt}=\langle a,b,c\rangle \cdot \langle x_2 - x_1, y_2 - y_1, z_2 - z_1 \rangle = a(x_2-x_1) + b(y_2-y_1) + c(z_2-z_1)$$ Since the force is constant, this expression does not depend on \(t\).

Compute the line integral

Now we are ready to compute the line integral: $$\int_{0}^{1} \mathbf{F} \cdot \frac{d\mathbf{r}(t)}{dt} \, dt = \int_{0}^{1} [ a(x_2-x_1) + b(y_2-y_1) + c(z_2-z_1) ] \, dt$$ This integral can be computed by multiplying the integrand by \(dt\) and integrating component-wise from \(0\) to \(1\). The result is: $$\int_{0}^{1} [ a(x_2-x_1) + b(y_2-y_1) + c(z_2-z_1) ] \, dt = a(x_2-x_1) + b(y_2-y_1) + c(z_2-z_1)$$

Compare the result to the dot product of the force and displacement vector

The expression above for the work done by the force is equal to the dot product of the force and the displacement vector \(\overrightarrow {AB}\): $$\mathbf{F} \cdot \overrightarrow{A B} = \langle a, b, c \rangle \cdot \langle x_2 - x_1, y_2 - y_1, z_2 - z_1 \rangle = a(x_2-x_1) + b(y_2-y_1) + c(z_2-z_1)$$ Thus, the work done in moving the object from point \(A\) to point \(B\) in the presence of a constant force \(\mathbf{F}\) is equal to \(\mathbf{F} \cdot \overrightarrow{A B}\) as required.

Key concepts

Constant Force and its Implications

When we talk about a constant force, we're referring to a force that remains the same in magnitude and direction throughout the movement of an object. This is important when calculating work, as it simplifies the mathematics involved. A constant force

• does not vary over time,
• ensures a uniform effect on the object,
• allows for straightforward calculations since the force vector \( \mathbf{F} = \langle a, b, c \rangle \) remains unchanged.

Imagine pushing a shopping cart with the same effort from start to end - that's a constant force! Understanding constant forces assists students in appreciating how these forces apply in real-world scenarios. Whether it's the gravity acting on a satellite or the thrust from a car engine, the concept remains the same.

Dot Product Fundamentals

The dot product is a way to multiply two vectors, resulting in a scalar. It is a core concept in physics, especially in the calculation of work. The dot product

• quantifies the component of one vector in the direction of another,
• is calculated as a combination of the magnitudes of the vectors and the cosine of the angle between them,
• for vectors \( \mathbf{F} = \langle a, b, c \rangle \) and \( \overrightarrow{AB} = \langle x_2-x_1, y_2-y_1, z_2-z_1 \rangle \), it becomes \( a(x_2-x_1) + b(y_2-y_1) + c(z_2-z_1) \).

The resulting scalar indicates how much of the force \( \mathbf{F} \) aligns with the direction of displacement \( \overrightarrow{AB} \). This calculation is critical in determining work, as the work done is the force along the path multiplied by the distance.

Path Parameterization: A Necessary Translation

Path parameterization translates a path into a mathematical form that is easier to manage. By parameterizing, we express the points along the path as functions of a single variable, often denoted as \(t\). For a straight-line path from point \(A\) \( \langle x_1, y_1, z_1 \rangle \) to point \(B\) \( \langle x_2, y_2, z_2 \rangle \), the parameterized path is \[ \mathbf{r}(t) = \langle x_1 + t(x_2-x_1), y_1 + t(y_2-y_1), z_1 + t(z_2-z_1) \rangle \].This expression

• captures the linear transition from point A to point B as \(t\) runs from 0 to 1,
• lets us compute derivatives needed for further calculations,
• simplifies the integration process in the work calculation.

This approach makes it easier to analyze any curve and is a valuable tool in both mathematics and physics.

The Mechanics of Work Calculation

Calculating work involves integrating the dot product of force and displacement over a path. Using the concept of a line integral, we derive how much work is done moving along this path in the presence of a force. In this context, work is expressed as \[ \int_{0}^{1} \mathbf{F} \cdot \frac{d\mathbf{r}(t)}{dt} \, dt \].Here's how it unfolds:

• Calculate the derivative of the parameterized path, \( \frac{d}{dt} \mathbf{r}(t) = \langle x_2 - x_1, y_2 - y_1, z_2 - z_1 \rangle \).
• Find the dot product of this derivative with constant force, resulting in a constant expression because the force doesn't change with \(t\).
• Integrate this constant expression over \(t\) from 0 to 1.

The computation confirms the formula for work done by a constant force as \( \mathbf{F} \cdot \overrightarrow{AB} \), representing the projection of force along the displacement - a foundational result in physics and engineering.