91Ó°ÊÓ

CALC An object has several forces acting on it. One force is \(\vec{\boldsymbol{F}}=\alpha x y \hat{\boldsymbol{r}},\) a force in the \(x\) -direction whose magnitude depends on the position of the object. (Sce Problem \(6.98 . )\) The constant is \(\alpha=2.00 \mathrm{N} / \mathrm{m}^{2}\) . The object moves along the following path: \((1)\) It starts at the origin and moves along the \(y\) -axis to the point \(x=0\) , \(y=1.50 \mathrm{m} ;(2)\) it moves parallel to the \(x\) -axis to the point \(x=1.50 \mathrm{m}, y=1.50 \mathrm{m} ;(3)\) it moves parallel to the \(y\) -axis to the point \(x=1.50 \mathrm{m}, y=0 ;(4)\) it moves parallel to the \(x\) -axis back to the origin. (a) Sketch this path in the \(x y\) -plane. (b) Calculate the work done on the object by \(\vec{F}\) for each leg of the path and for the complete round trip. (c) Is \(\vec{F}\) conservative or nonconservative? Explain.

Step by step solution

01

Sketch the Path in the xy-plane

First, draw a graph with the x-axis and y-axis. Plot the following points on this graph: the origin (0, 0), (0, 1.5), (1.5, 1.5), (1.5, 0). Connect these points to form a closed loop: a rectangle with the longer side along the y-axis. This represents the path of the object.

02

Calculate Work for Path Segment 1 (Origin to y=1.5)

For the movement along the y-axis (from (0,0) to (0,1.5)), the force \( \vec{F} \) has no component because it acts only in the x-direction (\( x=0 \) implies \( F = 0 \)). Thus, the work done \( W_1 = 0 \).

03

Calculate Work for Path Segment 2 (y=1.5 to x=1.5)

For movement along the x-axis from (0, 1.5) to (1.5, 1.5), at every point \( y = 1.5 \), and the force \( F = \alpha x y \). Calculate work as: \[ W_2 = \int_{0}^{1.5} F \cdot dx = \int_{0}^{1.5} \alpha x \cdot 1.5 \, dx = 1.5 \alpha \int_{0}^{1.5} x \, dx = 1.5 \alpha \left[\frac{x^2}{2}\right]_{0}^{1.5} \]Using \( \alpha = 2.00 \), this simplifies to:\[ W_2 = 1.5 \times 2.00 \times \frac{1.5^2}{2} = 3.375 \, \text{Joules} \]

04

Calculate Work for Path Segment 3 (x=1.5 from y=1.5 to y=0)

Along the vertical slide (from (1.5, 1.5) to (1.5, 0)), the force \( \vec{F} \) again contributes no work because it acts in the x-direction (no movement in x); thus, \( W_3 = 0 \).

05

Calculate Work for Path Segment 4 (x=1.5 to Origin)

Returning along the x-axis to the origin from (1.5, 0) to (0, 0) has \( y = 0 \), which results in force \( \vec{F} = 0 \). Therefore, the work done \( W_4 = 0 \).

06

Calculate Total Work for the Complete Round Trip

Sum up work done on all segments: \[ W_{total} = W_1 + W_2 + W_3 + W_4 = 0 + 3.375 + 0 + 0 = 3.375 \, \text{Joules} \]

07

Determine if the Force is Conservative or Nonconservative

Since the total work done is not zero \((3.375 \, \text{Joules})\) after a round trip path that ends where it started, the force \( \vec{F} \) is nonconservative.

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Key Concepts

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

Path Integral

A path integral is an essential concept in physics used to compute quantities like work or energy over a particular path in space. It involves integrating a quantity, typically related to forces or fields, over a specified path.
In the given problem, the path integral is used to calculate the work done by the force \( \vec{F} = \alpha x y \hat{\boldsymbol{r}} \) along a specified path. This path forms a rectangle on the xy-plane. The process involves breaking down the movement into different segments and calculating the work for each segment.
The total work is obtained by summing the work done on each segment. In our case, the work on horizontal segments along the x-axis and zero work for the vertical segments determine the total work done. Understanding path integrals is crucial as it allows us to find how much energy is required or used to move an object along a specific trajectory.

Conservative Forces

Conservative forces are forces where the work done moving along a closed path is zero. These forces have a potential energy associated with them, such as gravitational or electrostatic forces.
To determine if a force is conservative, check if the work done along any closed path is zero. For example, gravitational force acting on a heavy object does not depend on the path it takes.
In the given problem, the work done along the closed path was calculated to be 3.375 Joules instead of zero, thus the force \( \vec{F} \) is not conservative. This indicates that there's an energy loss or gain when returning to the starting point, differing from the path-independent nature of conservative forces.

Nonconservative Forces

Nonconservative forces, unlike conservative forces, do not have a definite relation to potential energy. They are path-dependent, meaning the work done by these forces depends on the path taken.
Examples of nonconservative forces include friction, air resistance, and the force provided by engines or brakes in vehicles.
In this problem, the force \( \vec{F} = \alpha x y \hat{\boldsymbol{r}} \) was determined to be nonconservative, as the total work done around the closed path was 3.375 Joules. This means energy could be lost to other forms, like heat, over this cycle, and path significantly influences the total work done.

Force Fields

Force fields are a way to describe how forces like gravity or electrostatics are distributed in space. Each point within the field has a force vector indicating the direction and magnitude of the force acting on an object at that point.
In this problem, we dealt with a force field given by \( \vec{F} = \alpha x y \hat{\boldsymbol{r}} \). The variable nature of this field means that the force's strength changes based on position in the xy-plane.
This field variation plays a critical role in determining the work done as the object navigates through different points, highlighting how certain areas have stronger or weaker forces depending on their field position. Understanding force fields aids in predicting the motion, stability, or energy requirements of objects subjected to such forces.