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Magazine Chapter 7: Problem 35
The force on a particle is directed along an \(x\) axis and given by \(F=F_{0}\left(x / x_{0}-1\right)\). Find the work done by the force in moving the particle from \(x=0\) to \(x=2 x_{0}\) by (a) plotting \(F(x)\) and measuring the work from the graph and (b) integrating \(F(x)\).
Short Answer
Expert verified
The work done by the force is 0 J.
Step by step solution
01
Understanding the Problem
We are tasked with finding the work done by a force on a particle moving along the x-axis, described by the function \( F = F_0 \left( \frac{x}{x_0} - 1 \right) \). The goal is to compute the work from \(x = 0\) to \(x = 2x_0\).
02
Analyzing the Force Function
The given force function, \(F(x) = F_0 \left( \frac{x}{x_0} - 1 \right)\), is linear in \(x\). At \(x = 0\), \(F = -F_0\), and at \(x = 2x_0\), \(F = F_0\). This indicates a linear increase in force from \(-F_0\) to \(+F_0\) over the interval \([0, 2x_0]\).
03
Plotting the Force Function
To visualize the work, we plot \(F(x)\). The function is a straight line with a negative slope starting at \(-F_0\) when \(x = 0\) and reaching \(F_0\) at \(x = 2x_0\). The x-intercept is at \(x = x_0\).
04
Calculating Work Using the Graph
The work done by a force is the area under the force vs. position graph. This area forms a trapezoid with bases \(F_0\) and \(-F_0\), and height \(2x_0\). The area (work) can be computed as \( \frac{1}{2} \times (F_0 + (-F_0)) \times 2x_0 = 0 \, \text{J} \).
05
Setting Up the Integral for Work
Work done by the force can also be calculated by integrating \(F(x)\) with respect to \(x\) over the interval \([0, 2x_0]\). The integral is: \( W = \int_{0}^{2x_0} F_0 \left( \frac{x}{x_0} - 1 \right) dx \).
06
Solving the Integral
Evaluate the integral to find the work: \[ W = F_0 \int_{0}^{2x_0} \left( \frac{x}{x_0} - 1 \right) dx = F_0 \left( \frac{1}{2x_0} x^2 - x \right) \Bigg|_{0}^{2x_0}. \] Substituting the limits, we get: \[ W = F_0 \left( \frac{1}{2x_0}(4x_0^2) - 2x_0 \right) - F_0 (0) = F_0 \left( 2x_0 - 2x_0 \right) = 0 \, \text{J}. \]
07
Conclusion
Both graphical and analytical methods confirm that the work done by the force when moving the particle from \(x = 0\) to \(x = 2x_0\) is zero, indicating that the force is fully balanced across this interval.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Force Function
A force function describes how force varies with respect to position, time, or another variable. In this exercise, we are examining a specific force function: \( F(x) = F_0 \left( \frac{x}{x_0} - 1 \right) \). This function indicates that force is linearly dependent on the position \( x \) along the x-axis.
When \( x = 0 \), the value of the force is \( -F_0 \). At the other extreme, when \( x = 2x_0 \), the value of the force is \( F_0 \). This means the force increases directly in proportion to the position of the particle, illustrating a linear change in force as you move along the x-axis.
When \( x = 0 \), the value of the force is \( -F_0 \). At the other extreme, when \( x = 2x_0 \), the value of the force is \( F_0 \). This means the force increases directly in proportion to the position of the particle, illustrating a linear change in force as you move along the x-axis.
- Linear Force: The function shows that the force starts negative and grows to positive, creating a balance in opposing directions over the motion range \([0, 2x_0]\).
- Visualization: Plotting \( F(x) \) results in a straight line graph making it easy to analyze the behaviour of the force applied to the particle.
Integrating Force
To calculate the work done by a force, we often need to integrate the force with respect to position. This is numerically equivalent to finding the area under a force versus position graph.
In mathematical terms, integrating the specific force function here requires the calculation:\[W = \int_{0}^{2x_0} F_0 \left( \frac{x}{x_0} - 1 \right) dx\]
In mathematical terms, integrating the specific force function here requires the calculation:\[W = \int_{0}^{2x_0} F_0 \left( \frac{x}{x_0} - 1 \right) dx\]
- Setting up the Integral: This equation asks us to find the integral of our force function over the interval from \( x = 0 \) to \( x = 2x_0 \). It represents a more rigorous way to determine the work done than simply interpreting the graph.
- Carrying out the Integration: As calculated, this yields \( 0 \). This result reaffirms the graph-based solution, as the positive and negative areas under the plotted line cancel each other out.
Work-Energy Principle
The work-energy principle connects the concepts of work and energy, stating that work done on an object is equivalent to the change in its kinetic energy. In simpler terms, when a force does work on a particle, it can change the particle's speed or position.
In this specific exercise, when examining the interval from \( x = 0 \) to \( x = 2x_0 \), we see that the total work done by the force comes out to be zero. But, why? The nature of our force function plays a crucial role:
In this specific exercise, when examining the interval from \( x = 0 \) to \( x = 2x_0 \), we see that the total work done by the force comes out to be zero. But, why? The nature of our force function plays a crucial role:
- Opposing Forces: Since the force transitions from negative to positive, the work done in one half of the motion cancels out the work done in the other half.
- No Net Energy Change: As there is zero net work, there is also no net change in kinetic energy, meaning the particle has not changed in terms of its speed by moving from its initial to the final position within the given bounds.
