Problem 98 - University Physics with Modern Physics

Question

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

Step-by-Step Solution

Answer

(a) 15 J, (b) 0 J, (c) 10 J.

Step 1: Understanding Force Equation

The given force is \( \vec{\boldsymbol{F}} = \alpha x y \hat{\boldsymbol{r}} \). Since \( \alpha = 2.50 \, \text{N/m}^2 \), the force can be expressed as \( \vec{\boldsymbol{F}} = 2.5 x y \hat{\boldsymbol{r}} \). This indicates that the force's magnitude depends on the coordinates \(x\) and \(y\).

Step 2: Work Done in Displacement (a) - Along x-axis

For displacement along the x-axis, the displacement vector is \( dx \hat{\boldsymbol{r}} \). The work done \( W \) by the force is the integral of the force along the path of motion. The force is \( F = 2.5xy \) and the displacement is from \( x=0 \) to \( x=2 \), with \( y \) constant at \( 3 \). The work is \( W = \int_{0}^{2} 2.5xy \, dx = 2.5 \times 3 \int_{0}^{2} x \, dx \). Solving the integral, \( W = 2.5 \times 3 [\frac{x^2}{2}]_{0}^{2} = 2.5 \times 3 \times 2 = 15 \, \text{J} \).

Step 3: Work Done in Displacement (b) - Along y-axis

For displacement along the y-axis from \( y=0 \) to \( y=3 \) with \( x=2 \), no work is done because the force is along the x-direction and displacement is along the y-direction. Therefore, \( W = 0 \).

Step 4: Work Done in Displacement (c) - Along y=1.5x

For the displacement along \( y = 1.5x \), substitute \( y = 1.5x \) into \( F = 2.5xy \). Then, \( F = 2.5x(1.5x) = 3.75x^2 \). The work done is calculated by integrating \( F \, dx \) from \( x=0 \) to \( x=2 \). Thus, \( W = \int_{0}^{2} 3.75x^2 \, dx = 3.75 \int_{0}^{2} x^2 \, dx = 3.75 [\frac{x^3}{3}]_{0}^{2} = 3.75 \times \frac{8}{3} = 10 \, \text{J} \).

Key Concepts

Displacement Along Axis

When analyzing the work done by a particular force, understanding the concept of displacement along an axis is crucial. Displacement refers to the change in position of an object. It can occur in different directions such as parallel to the x-axis or y-axis, or along any path. Each case influences the calculation of work done by force differently.

In displacement parallel to the x-axis, like in scenario (a) from the exercise, the displacement direction aligns with the force's action, meaning work is performed as the object moves. Conversely, for movement purely along the y-axis, like in scenario (b), there's no work done by a force that only acts in the x-direction.

For movement along a path that's neither purely x nor y, like line \(y = 1.5x\) in scenario (c), both components of the position \(x\) and \(y\) play a role. Here, the path function is substituted into the force equation to evaluate the work done accurately.

Integral of Force

The integration of force along a path is needed to calculate the work done. In physics, work is defined as the integral of force over a displacement path. This essentially means we accumulate the effect of the force as the object moves.

For motion along a single axis where the force also acts, the path of integration is straightforward. For example, in displacement (a), the integral \(\int_{0}^{2} 2.5xy \, dx\) simplifies since \(y\) is constant. Only \(x\) varies, focusing integration on \(2.5 \times 3 \int_{0}^{2} x \, dx\), resulting in a direct calculation of the area under the curve.

However, when movement occurs through varied paths such as lines like \(y=1.5x\), integration becomes more complex. We substitute \(y = 1.5x\), transforming the integral into \(3.75 \int_{0}^{2} x^2 \, dx\), thereby capturing exactly how force varies with position as both \(x\) and \(y\) vary.

Force Equation

The force equation describes how the force's magnitude depends on the object's position. In this exercise, the given force \(\vec{\boldsymbol{F}} = \alpha x y \hat{\boldsymbol{r}}\) shows a position-dependent force in the x-direction.

Breaking down the force equation, \(\alpha x y\) indicates that force varies with the product of \(x\) and \(y\). This multiplication means both coordinates directly affect the force's value. For instance, increasing either \(x\) or \(y\) increases the total force experienced by the object at that point along its path.

This force characteristic results in work differences across distinct paths. When \(y\) is held constant, like scenario (a), \(x\) solely determines the force's change. But in scenario (c) along \(y = 1.5x\), both \(x\) and \(y\) vary together, reshaping the force function to \(3.75x^2\), highlighting the inherent variability and making exact path-dependent work calculations possible.