Problem 75 - University Physics with Modern Physics

Question

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

Step-by-Step Solution

Answer

(a) -50.625 J, (b) -67.50 J, (c) \( \overrightarrow{F} \) is nonconservative.

Step 1: Define the Force and Path

The force acting on the tool is \( \overrightarrow{F} = -\alpha xy^2 \hat{\jmath} \), with \( \alpha = 2.50 \) N/m\(^3\). We need to calculate the work done when the tool travels along two different paths. First, calculate for the straight line path from the origin to point (3.00 m, 3.00 m) where \( y = x \).

Step 2: Calculate Work Along the Straight Line Path

The straight-line path is \( y = x \), thus the force simplifies to \( \overrightarrow{F} = -\alpha x^3 \hat{\jmath} \). The displacement vector \( \overrightarrow{ds} = dx \hat{i} + dy \hat{j} \) becomes \( dx \hat{i} + dx \hat{j} \) because \( dy = dx \) for \( y = x \).Calculate the work:\[W = \int_{0}^{3} \overrightarrow{F} \cdot \overrightarrow{ds} = \int_{0}^{3} -\alpha x^3 \hat{j} \cdot \hat{j} dx = -\alpha \int_{0}^{3} x^3 dx = -\alpha \left[ \frac{x^4}{4} \right]_{0}^{3} = -2.50 \left[ \frac{3^4}{4} \right] = -\frac{2.50 \cdot 81}{4} = -50.625 \, \text{J}.\]

Step 3: Calculate Work for Path Along Axes

First, the tool moves along the x-axis from (0, 0) to (3, 0). Along this path \( y = 0 \), so \( \overrightarrow{F} = 0 \). Therefore, \( W_x = 0 \) J.Next, it moves parallel to the y-axis from (3, 0) to (3, 3). Here, the force simplifies to \( \overrightarrow{F} = -\alpha \cdot 3 \cdot y^2 \hat{\jmath} \).Calculate the work:\[W_y = \int_{0}^{3} \overrightarrow{F} \cdot \overrightarrow{ds} = \int_{0}^{3} -7.50 \cdot y^2 \hat{j} \cdot dy = -7.50 \int_{0}^{3} y^2 dy = -7.50 \left[ \frac{y^3}{3} \right]_{0}^{3} = -7.50 \cdot \frac{27}{3} = -67.50 \, \text{J}.\]

Step 4: Compare Work and Determine Force Nature

The work along path 1 (straight line) is \(-50.625\) J, while along path 2 (via axes) it is \(-67.50\) J. Since the work done by the force differs depending on the path, this indicates that \( \overrightarrow{F} \) is nonconservative, as the work is path-dependent.

Key Concepts

Nonconservative Forces

In physics, forces can be categorized into conservative and nonconservative forces. A **conservative force** is one where the work done is independent of the path taken; such forces are only dependent on the initial and final states. **Nonconservative forces**, like the one in our exercise, do not have this property. This means that the work done by these forces depends on the path taken between two points.

In the given problem, the force \( \overrightarrow{F} = - \alpha xy^2 \hat{\jmath} \) acts on the tool in a direction opposite to the y-axis. Because the work done by this force varies when we adjust the path from one point to another, we confirm it is nonconservative. Such forces often involve friction or air resistance and even complex path-dependent forces like the ones described here.

Recognizing whether a force is nonconservative is crucial in work-energy principles since it directly impacts energy calculations and system analysis. This understanding helps predict the total work done on an object depending on the paths considered.

Path-dependent Work

In situations involving nonconservative forces, **path-dependent work** becomes a crucial concept. It refers to work calculations that vary based on the trajectory or route taken by an object.

Let's dissect the problem at hand. The tool moves along two distinct paths to reach the same final position. In the first scenario, it travels straight along the line **y = x**. The work done here is calculated to be \(-50.625 \, \text{J}\). However, when the tool follows a path via the x-axis to the point (3, 0) and then directly along the y-axis to (3, 3), the work increases to \(-67.50 \, \text{J}\).

This difference signifies the path dependency of the work done by \( \overrightarrow{F} \). Such forces usually require a more detailed analysis to ensure that the complete energy transformations are understood. Recognizing path dependency is fundamentally different from conservative systems that only consider initial and final energy states.

Force Calculations

**Force Calculations** are essential for determining work when forces act on an object over a distance. In our exercise, the force acting on a tool is expressed mathematically as a function of its coordinates, \( \overrightarrow{F} = - \alpha xy^2 \hat{\jmath} \). Understanding how to calculate the impact of such a force along different paths requires breaking down the movement into components.

During the straight-line movement along **y = x**, the force simplifies due to geometry. The integral formula for work allows us to find \( W = \int \overrightarrow{F} \cdot \overrightarrow{ds} \), calculating to \(-50.625 \, \text{J}\). For paths traversing the axes, the calculation must account for the changing influence of \( y^2 \) along segments where movement is purely in the x or y direction.

Successfully calculating these values requires a though understanding of vector calculus. It's important to compute the dot product \( \overrightarrow{F} \cdot \overrightarrow{ds} \) correctly to integrate over the path. Through these calculations, physics students grasp how forces transform across distances, which is central to work-energy principles.