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An object is attracted toward the origin with a force given by \(F_{x}=-k / x^{2} .\) (Gravitational and electrical forces have this distance dependence.) (a) Calculate the work done by the force \(F_{x}\) when the object moves in the \(x\) -direction from \(x_{1}\) to \(x_{2}\) . If \(x_{2}>x_{1}\) , is the work done by \(F_{x}\) positive or negative? (b) The only other force acting on the object is a force that you exert with your hand to move the object slowly from \(x_{1}\) to \(x_{2}\) . How much work do you do? If \(x_{2}>x_{1},\) is the work you do positive or negative? (c) Explain the similarities and differences between your answers to parts (a) and (b).

Short Answer

Expert verified
(a) Positive; (b) Negative; (c) Equal magnitude, opposite sign due to energy conservation.

Step by step solution

01

Understanding Force and Work

The force given is \( F_{x} = -\frac{k}{x^2} \), where it is dependent on the distance from the origin. The work \( W \) done by a force is calculated by integrating the force over the path of the object: \( W = \int_{x_1}^{x_2} F_{x} \, dx \).
02

Integrating the Force

Substitute the force equation into the work integral: \( W = \int_{x_1}^{x_2} -\frac{k}{x^2} \, dx \). The integral of \(-\frac{k}{x^2}\) with respect to \(x\) is \(\frac{k}{x}\). Thus, \( W = \left[-\frac{k}{x}\right]_{x_1}^{x_2} \).
03

Evaluating the Integral

Evaluate the integral from \(x_1\) to \(x_2\): \[ W = \left( -\frac{k}{x_2} + \frac{k}{x_1} \right) \]. Simplifying, the work done by the force is \( W = -k \left( \frac{1}{x_2} - \frac{1}{x_1} \right) \).
04

Determining Sign of Work

Since \( x_2 > x_1 \), \( \frac{1}{x_2} < \frac{1}{x_1} \), making \( \frac{1}{x_2} - \frac{1}{x_1} \) negative. Therefore, the work done by the force \( F_x \) is positive: \( k \left( \frac{1}{x_1} - \frac{1}{x_2} \right) \).
05

Calculating Work by Hand Force

The hand's force must counteract \( F_{x} \) to move the object slowly, implying it does work that is equal in magnitude but opposite in sign. Hence, the work done by your hand is \( -W = k \left( \frac{1}{x_2} - \frac{1}{x_1} \right) \), which is negative since \( x_2 > x_1 \).
06

Comparing both Works

For part (a), the force \( F_x \) does work (positive) in moving the object. For part (b), your hand does work (negative) to maintain slow movement. The comparison shows that they are equal in magnitude but opposite in sign, reflecting energy conservation principles.

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Key Concepts

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

Gravitational Force
Gravitational force is a fundamental force of nature that governs the attraction between masses. It's a force that one object exerts on another object because of their masses. One key characteristic of gravitational forces, as shown in this exercise, is that they follow an inverse square law. This means that the force decreases with the square of the distance between the two objects. Mathematically, it is represented as:\[F = -\frac{k}{x^2}\]where:
  • \( F \) is the force acting on the object,
  • \( k \) is a constant that depends on the specific scenario, such as the mass of the objects or other relevant factors,
  • \( x \) is the distance from the origin or object to the point where the force is being measured.
This relationship implies that as an object moves further away, the gravitational force exerted on it becomes weaker. Such understanding is crucial in fields like astrophysics and engineering because it impacts how objects in space or within gravitational fields behave and interact.
Force Integration
The concept of integrating a force over a distance is using calculus to find the total work done by this force as an object moves from one point to another. When you need to calculate work performed by a varying force like the one given here, you use integration to sum up the infinitesimal contributions over a path.In this exercise, the force given is \( F_x = -\frac{k}{x^2} \). To calculate the work \( W \), integrate this force along the path from \( x_1 \) to \( x_2 \) like so:\[W = \int_{x_1}^{x_2} F_x \, dx = \int_{x_1}^{x_2} -\frac{k}{x^2} \, dx\]The integration process for \(-\frac{k}{x^2}\) with respect to \( x \) results in \( \frac{k}{x} \). After evaluating from \( x_1 \) to \( x_2 \), the final expression for work is:\[W = -k \left( \frac{1}{x_2} - \frac{1}{x_1} \right)\]This integral shows how the changing force over distance results in total work being done. Understanding this concept helps in tailoring solutions for real-world applications where forces aren't constant, such as designing machines or calculating planetary motion.
Energy Conservation
The principle of energy conservation is crucial when analyzing forces and work. It states that the total energy in a closed system remains constant over time. Energy cannot be created or destroyed, only transferred or converted from one form to another.In this problem, energy conservation manifests in the work done by different forces. When the given force \( F_x \) does work, it is transferring energy to the object. Meanwhile, when slowly moving the object against this force with your hand, you are also exerting force that requires work.The work-energy principle explains that how work done by different forces must either store energy in or remove energy from the object. As the solution shows, the specified force does positive work (indicating energy transfer to the object), while the hand does negative work, exactly counterbalancing this energy transfer and maintaining slow, controlled motion. Consequently, no net change in the energy of the system confirms the energy conservation law.Understanding energy conservation not only elucidates the behavior of systems in equilibrium but also guides engineers and physicists in predicting how different forces interact within various scenarios.

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Most popular questions from this chapter

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