Problem 83 - University Physics with Modern Physics

Question

A cutting tool under microprocessor control has several forces acting on it. One force is \(\overrightarrow{\boldsymbol{F}}=-\alpha x y^{2} \hat{\boldsymbol{j}},\) a force in the negative \(y\) -direction whose magnitude depends on the position of the tool. The constant is \(\alpha=2.50 \mathrm{N} / \mathrm{m}^{3} .\) Consider the displacement of the tool from the origin to the point \(x=3.00 \mathrm{m}, y=3.00 \mathrm{m} .\) (a) Calculate the work done on the tool by \(\vec{F}\) if this displacement is along the straight line \(y=x\) that connects, these two points. (b) Calculate the work done on the tool by \(\vec{F}\) if the tool is first moved out along the \(x\) -axis to the point \(x=3.00 \mathrm{m}, y=0\) and then moved parallel to the \(y\) -axis to the point \(x=3.00 \mathrm{m}\) , \(y=3.00 \mathrm{m}\) . (c) Compare the work done by \(\vec{F}\) along these two paths. Is \(\vec{F}\) conservative or nonconservative? Explain.

Step-by-Step Solution

Answer

Work is -50.63 J along y=x; -67.5 J along axis path. The force is nonconservative.

Step 1: Define the Force Field

The force acting on the tool is given as \( \overrightarrow{\boldsymbol{F}} = -\alpha x y^2 \hat{\boldsymbol{j}} \), and it acts in the negative \( y \)-direction. The constant \( \alpha \) is 2.50 N/m³. This force depends on the position \( (x, y) \) of the tool.

Step 2: Express Work Formula

Work \( W \) done by a force is given by the integral of the force along the path of displacement: \( W = \int \overrightarrow{\boldsymbol{F}} \cdot d\overrightarrow{\boldsymbol{r}} \). For a path from point A to point B, \( d\overrightarrow{\boldsymbol{r}} \) is your line element and changes depending on the path taken.

Step 3: Path along y = x

Along the path \( y = x \), the line element \( d\overrightarrow{\boldsymbol{r}} = dx \hat{\boldsymbol{i}} + dy \hat{\boldsymbol{j}} \). Since \( y = x \), we have \( dy = dx \). The force then becomes \( \overrightarrow{\boldsymbol{F}} = -\alpha x x^2 \hat{\boldsymbol{j}} = -\alpha x^3 \hat{\boldsymbol{j}} \). Integrate from \( x = 0 \) to \( x = 3 \). Work \( W = \int_0^3 (-\alpha x^3) dx = -\alpha \int_0^3 x^3 dx \).

Step 4: Calculate Work for y = x

The integral \( \int_0^3 x^3 dx = \left[ \frac{x^4}{4} \right]_0^3 = \frac{81}{4} \). Thus, \( W = -2.5 \times \frac{81}{4} = -50.625 \text{ J} \).

Step 5: Path along the x-axis and y-axis

For the first segment (\( y = 0 \)), \( \,d\overrightarrow{\boldsymbol{r}} = dx \hat{\boldsymbol{i}} \), and \( F_j = 0 \), so \( W_1 = 0 \). For the second segment, keep \( x \) constant at 3 m and integrate \( \overrightarrow{\boldsymbol{F}} = -\alpha \, 3 \, y^2 \hat{\boldsymbol{j}} \, \): \( W_2 = \int_0^3 (-\alpha \, 3 \, y^2) \, dy \).

Step 6: Calculate Work along Axes Path

Compute the integral: \( \int_0^3 y^2 \, dy = \left[ \frac{y^3}{3} \right]_0^3 = 9 \). Thus, \( W_2 = -3 \, \times \, 2.5 \, \times \, 9 = -67.5 \, \text{J} \). The total work for this path is \( W_1 + W_2 = 0 - 67.5 = -67.5 \, \text{J} \).

Step 7: Compare the Work Done

The work done along the two paths is different: \(-50.625 \text{ J}\) along \( y = x \) and \(-67.5 \text{ J}\) when moving along the \( x \)-axis then \( y \)-axis. Since the work depends on the path taken, the force \( \overrightarrow{\boldsymbol{F}} \) is nonconservative.

Key Concepts

Conservative and Nonconservative Forces

In physics, forces are categorized as either conservative or nonconservative based on how they perform work along different paths. A conservative force is one where the work done by the force on an object moving from one point to another is path-independent. Examples of conservative forces include gravitational and electrostatic forces.

Nonconservative forces, on the other hand, depend on the path taken between the initial and final points. In the original exercise, we identified the force as nonconservative because the work done varied based on the path followed. We calculated different work values for a straight path and a two-segment path. This indicates that the force cannot be described by a potential energy, characteristic of nonconservative forces like friction.

Understanding whether a force is conservative or nonconservative is crucial. It determines if we can apply energy conservation principles directly or need to account for energy dissipation or transformation in analyses.

Path Integrals

Path integrals are a way to calculate the work done by a force over a certain path. When dealing with variable forces, the simple formula of force times distance is not sufficient. Instead, we integrate the force over the path the object travels.

This integration considers all points along the path, taking the dot product between the force vector and the infinitesimal displacement vector. In the exercise, this was crucial for determining the work done by using the formula:
\[ W = \int_{C} \overrightarrow{F} \cdot d\overrightarrow{r} \]
Where \( C \) is the path taken. Due to directional dependencies and variations in force magnitude, the path integral allowed for an accurate calculation of work.

• For different paths, the configuration of path integrals changes.
• The integral's path dependence highlights the nature of the force (conservative or nonconservative).

Path integrals are essential in understanding complex systems where forces change across positions.

Vector Calculus

Vector calculus is indispensable for understanding physical phenomena involving multiple dimensions, like forces acting in different directions, as seen in our exercise. It allows us to handle vector fields, perform operations such as calculating work, and delve into concepts like gradients, curls, and divergences.

Vector calculus provides tools such as:

• **Gradient**: Describes the rate and direction of change in a field.
• **Divergence**: Measures the magnitude of a source or sink at a given point in a vector field.
• **Curl**: Indicates the rotation or circular motion within a vector field.

In the exercise, the force with components changing with position required us to calculate work, necessitating the use of vector calculus techniques. The differential line element \( d\overrightarrow{r} \) included both \( dx \hat{i} \) and \( dy \hat{j} \), showcasing how we break down paths into manageable vector components.

Understanding vector calculus is critical, as it bridges the gap between abstract mathematical operations and physical applications in fields like electromagnetism, fluid dynamics, and engineering.