91Ó°ÊÓ

CALC A cutting tool under microprocessor control has several forces acting on it. One force is \(\vec{\boldsymbol{F}}=-\alpha x y^{2} \hat{\boldsymbol{J}}\) , 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 \(\vec{\boldsymbol{F}}\) if this displacement is along the straight line \(y=x\) that connects these two points. (b) Calculate the work done on the tool by \(\vec{\boldsymbol{F}}\) 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 \(\vec{\boldsymbol{F}}\) along these two paths. Is \(\vec{\boldsymbol{F}}\) conservative or nonconservative? Explain.

Step by step solution

01

Understand the Problem

We need to calculate the work done by the force \(\vec{\boldsymbol{F}} = -\alpha x y^2 \hat{\boldsymbol{J}}\) over two different paths from the origin to the point \((3, 3)\). The force depends on the position \((x, y)\) and acts in the negative \(y\)-direction. The constant \(\alpha = 2.50 \ \text{N/m}^3\).

02

Calculate Work for Path (a)

For path (a), the displacement is along the line \(y = x\), so we can write the force as \(\vec{\boldsymbol{F}} = -\alpha x (x^2) \hat{\boldsymbol{J}} = -\alpha x^3 \hat{\boldsymbol{J}}\). The work done is given by the integral \(W = \int \vec{\boldsymbol{F}} \cdot d\vec{r}\). Along this path, \(d\vec{r} = dx \hat{\boldsymbol{I}} + dy \hat{\boldsymbol{J}}\), and \(dy = dx\) (since \(y=x\)). Hence, \(d\vec{r} = dx\hat{\boldsymbol{I}} + dx\hat{\boldsymbol{J}}\). The dot product \(\vec{\boldsymbol{F}} \cdot d\vec{r} = -\alpha x^3 dx\). Therefore, the work is \[W = \int_0^3 -\alpha x^3 \, dx = -\alpha \int_0^3 x^3 \, dx = -\frac{\alpha x^4}{4} \Bigg|_0^3\]. Substitute \(\alpha = 2.50\) and evaluate to get \[W = -\frac{2.5 \times 3^4}{4} = -50.625 \ \text{J} \].

03

Calculate Work for Path (b)

For path (b), first move along the \(x\)-axis from \((0,0)\) to \((3,0)\), where \(y=0\) and the force is zero. Therefore, the work done is \(0 \text{ J}\). Next, move parallel to the \(y\)-axis from \((3,0)\) to \((3,3)\), along which \(x=3\). The force becomes \(\vec{\boldsymbol{F}} = -\alpha \times 3 \times y^2 \hat{\boldsymbol{J}} = -7.5 y^2 \hat{\boldsymbol{J}}\). The displacement in the \(y\) direction is \(dy\), so the dot product is \(-7.5 y^2 \, dy\). Thus, \[W = \int_0^3 -7.5 y^2 \, dy = -7.5 \frac{y^3}{3} \Bigg|_0^3 = -67.5 \ \text{J} \].

04

Compare Work for Both Paths and Assess Conservativeness

For path (a), the work done is \(-50.625 \ \text{J}\), and for path (b), it is \(-67.5 \ \text{J}\). Since the work done by \(\vec{\boldsymbol{F}}\) depends on the path taken, \(\vec{\boldsymbol{F}}\) is a nonconservative force.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Nonconservative Forces

Nonconservative forces are forces for which the work done depends on the path taken, rather than just the initial and final positions. These forces, like friction or air resistance, typically involve energy being transformed into other forms, like heat, which cannot be easily recovered. In the exercise above, the force \( \vec{\boldsymbol{F}} = -\alpha x y^2 \hat{\boldsymbol{J}} \) is shown to be nonconservative. The key evidence is that the work done by this force on the tool is different when the tool follows different paths between the same two points.

• Path (a): work done = \-50.625\ J
• Path (b): work done = \-67.5\ J

Since the calculated work is not the same for both paths, it highlights the characteristic of nonconservative forces.

Vector Calculus

Vector calculus involves mathematical tools to analyze and understand the forces and movements in physics. Concepts like vectors, line integrals, and gradients are integral to this field. In the given problem, vector calculus is employed through the vector \(\vec{\boldsymbol{F}} = -\alpha x y^2 \hat{\boldsymbol{J}}\) to calculate the work done. The vector representation helps in breaking down the problem into components, which makes the integration more straightforward. By understanding the vector

• Direction and magnitude: \( \hat{\boldsymbol{J}} \) indicates the force acts in the \(y\)-direction.
• Dependence on position: Highlighted by terms like \(x y^2\).

This field allows for precise calculations that consider the variations at every point in a given path.

Path Independence

Path independence is a property that belongs to conservative forces. It means that the work done by the force between two points is the same regardless of the path taken. This is not the case here, demonstrating the concept of path dependence in nonconservative forces. With the tool's movement

• Along the straight line \(y=x\) results in work of \(-50.625\ J\).
• Along the \(x\)-axis, then \(y\)-axis results in work of \(-67.5\ J\).

These different results confirm the absence of path independence for \(\vec{\boldsymbol{F}}\). In contrast, systems where path independence occurs, such as gravitational or electrostatic fields, would show no variation in work due to the path chosen.

Line Integrals

A line integral is a fundamental concept in vector calculus, used to calculate the work done by a force field along a path. In the exercise, line integrals help calculate the work done by \(\vec{\boldsymbol{F}}\) as the tool moves from one point to another. The line integral sums up the values of the force along the entire trajectory.

• For path (a), this becomes \( W = \int_0^3 -\alpha x^3 \, dx \).
• For path (b), the line integral splits into two parts since the movement is in distinct segments.

Using line integrals allows us to capture the varying nature of the force, especially when considering paths of differing shapes or segments. This approach provides a more comprehensive understanding of how a force interacts over the extent of the chosen path.