Question

An object has several forces acting on it. One of these forces is \(\overrightarrow{F}= axy\hat{\imath}\), a force in the \(x\)-direction whose magnitude depends on the position of the object, with \(\alpha = 2.50 \, \mathrm{N/m}^2\). Calculate the work done on the object by this force for the following displacements of the object: (a) The object starts at the point (\(x = 0\), \(y = 3.00\) m) and moves parallel to the x-axis to the point (\(x= 2.00\) m, \(y = 3.00\) m). (b) The object starts at the point (\(x = 2.00\) m, \(y = 0\)) and moves in the \(y\)-direction to the point (\(x = 2.00\) m, \(y = 3.00\) m). (c) The object starts at the origin and moves on the line \(y = 1.5x\) to the point (\(x = 2.00\) m, \(y = 3.00\) m).

Short answer

Path (a): 15 N·m, Path (b): 0 N·m, Path (c): 10 N·m.

Step-by-step solution

Identify Known Information

We are given the force function \( \overrightarrow{F} = \alpha x y \hat{\imath} \) with \( \alpha = 2.50 \, \mathrm{N/m}^2 \). We need to calculate the work done for three different paths. The positions and paths for each case are provided.

Path a: Integrate Force over Path

In path (a), the object moves from \((x = 0, y = 3.00 \text{ m})\) to \((x = 2.00 \text{ m}, y = 3.00 \text{ m})\) along the x-axis. The force in the x-direction is constant for \(y = 3\) m.\[ W_a = \int_{0}^{2.00} F_x \, dx = \int_{0}^{2.00} \alpha x (3) \, dx \]Substitute \( \alpha = 2.50 \, \mathrm{N/m}^2 \):\[ W_a = 3 \alpha \int_{0}^{2.00} x \, dx = 3(2.50) \left[ \frac{x^2}{2} \right]_0^{2.00} = 7.5 \left[ \frac{4}{2} \right] = 15 \, \text{N} \cdot \text{m} \]

Path b: Check Direction of Motion

In path (b), the object moves from \((x = 2.00 \text{ m}, y = 0)\) to \((x = 2.00 \text{ m}, y = 3.00 \text{ m})\) in the y-direction. Since the displacement is parallel to the y-axis, the force, which acts in the x-direction, does no work.\[ W_b = 0 \, \text{N} \cdot \text{m} \]

Path c: Define Path and Calculate

For path (c), the object moves from the origin along the line \(y = 1.5x\) to \((x = 2.00 \text{ m}, y = 3.00 \text{ m})\). Substitute \(y = 1.5x\) into the force equation:\[ \overrightarrow{F} = \alpha x (1.5x) \hat{\imath} = 1.5 \alpha x^2 \hat{\imath} \]The work done is:\[ W_c = \int_{0}^{2.00} 1.5 \alpha x^2 \, dx = 1.5(2.50) \int_{0}^{2.00} x^2 \, dx \]\[ W_c = 3.75 \left[ \frac{x^3}{3} \right]_0^{2.00} = 3.75 \left[ \frac{8}{3} \right] = 10 \, \text{N} \cdot \text{m} \]

Summarize Work Done for Each Path

- For path (a), the work done is \(15 \, \text{N} \cdot \text{m}\).- For path (b), the work done is \(0 \, \text{N} \cdot \text{m}\).- For path (c), the work done is \(10 \, \text{N} \cdot \text{m}\).

Key concepts

Integral Calculus in Physics

Integral calculus plays an essential role in understanding how forces act on objects and how work is performed. In physics, work is calculated by integrating the force applied to an object along a path. Let's break this down step by step:
- **Function of Force**: The force acting on the object is given as \( \overrightarrow{F} = \alpha x y \hat{\imath} \), where \( \alpha \) is a constant value. The force changes depending on the object's position.- **Integration Process**: To find the work done along a path, we integrate the force over that path. For example, in part (a) of the original problem, we integrated \( \alpha x (3) \) from \( x = 0 \) to \( x = 2.00 \) m. The process involved computing the area under the force curve.By using integration, we account for the variability in the force's magnitude and calculate the precise work done over the displaced distance. Framed as a tip, always remember that the limits of integration should match the displacement's starting and ending points.

Force and Motion

The relationship between force and motion is central to physics. In understanding this relationship, one must grasp a few foundational principles:
- **Force Direction**: In our problem, \( \overrightarrow{F} = \alpha x y \hat{\imath} \) indicates the force acts in the x-direction.- **Path Independence**: As evident in path (b), the force doesn't perform work if the motion is perpendicular to the force's direction. Even if the object moves from \( y=0 \) to \( y=3.00 \) m, as long as the displacement is along the y-axis (where \( x \) remains constant), no work is done.Understanding the vector nature of force is crucial. Forces have both magnitude and direction. Thus, they only do work when there's movement in their direction. This concept helps determine when work is being done and guides us in calculating it accurately.

Path-dependent Work Calculations

When calculating work, the path taken by the object can significantly affect the total work done. This is known as path-dependent calculations:
- **Different Paths, Different Work**: Paths (a), (b), and (c) in this problem highlight this variation. Despite starting and ending at the same points, different paths result in different amounts of work.- **Example Path (c)**: This illustrates the complexity of path-dependent work. As the object moves from the origin to another point along \( y = 1.5x \), we're required to substitute \( y = 1.5x \) into our force expression. Here, \( \overrightarrow{F} = 1.5 \alpha x^2 \hat{\imath} \), leading us to integrate \( 1.5 \alpha x^2 \) from \( x = 0 \) to \( x = 2.00 \) m.Ultimately, choosing different paths over which to perform integrations demonstrates how changes in path affect how much work is calculated. This intricacy means that each potential path must undergo its own calculation to understand the work performed.