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Chapter 6: Problem 74

A force in the \(+x\) -direction has magnitude \(F=b / x^{n}\) , where \(b\) and \(n\) are constants. (a) For \(n>1\) , calculate the work done on a particle by this force when the particle moves along the \(x\) -axis from \(x=x_{0}\) to infinity. (b) Show that for \(0< n< 1,\) even though \(F\) becomes zero as \(x\) becomes very large, an infinite amount of work is done by \(F\) when the particle moves from \(x=x_{0}\) to infinity.

Short Answer

Expert verified
(a) Finite work for \( n > 1 \), \(-\frac{b}{1-n} x_0^{1-n}\). (b) Infinite work for \( 0 < n < 1 \).

Step by step solution

01

Understanding Work Done by a Force

The work done by a force as a particle moves from one point to another is calculated using the integral of the force with respect to displacement. When the force is variable, we express it as an integral, \( W = \int_{x_0}^{\infty} F(x) \, dx \).
02

Expressing the Force

Given that the force is expressed as \( F(x) = \frac{b}{x^n} \), substitute this into the work integral to express the work done as \( W = \int_{x_0}^{\infty} \frac{b}{x^n} \, dx \).
03

Solving the Integral for n > 1

When solving the integral for \( n > 1 \), we perform the integration: \( \int \frac{b}{x^n} \, dx = \frac{b}{1-n} x^{1-n} + C \). Evaluating from \( x_0 \) to \( \infty \) gives \( W = \lim_{x \to \infty} \frac{b}{1-n} x^{1-n} - \frac{b}{1-n} x_0^{1-n} \). As \( x \to \infty \), assuming \( n > 1 \), \( x^{1-n} \to 0 \) and hence \( W = -\frac{b}{1-n} x_0^{1-n} \) which is a finite value.
04

Analyzing the Integral for 0 < n < 1

Now for \( 0 < n < 1 \), the setup remains the same, but integrating \( \int \frac{b}{x^n} \, dx = \frac{b}{1-n} x^{1-n} + C \) gives a different result. As \( x \to \infty \), \( x^{1-n} \to \infty \) which indicates that \( W = \infty \). Hence, an infinite amount of work is done.

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Key Concepts

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

Work-energy principle
The work-energy principle is a significant concept in physics, especially in the study of dynamics. It states that the work done by all forces acting on a particle is equal to the change in the kinetic energy of the particle.
This principle helps us understand how forces affect motion, and it can be used to calculate the work done when a force, especially a variable force, is applied over a distance.
In simpler terms, when a force is applied to an object to move it, the object gains or loses kinetic energy depending on the amount of work done by the force.
Work is positive when the force acts in the direction of motion and negative when it opposes the motion. For our exercise, the work done by a force is computed by considering how that force acts over a particular path. This involves using calculus to integrate the force over the distance traveled.
This integration gives us the total work done and connects with changes in energy resulting from the motion of the particle.
Definite Integrals
Definite integrals are a fundamental part of calculus used to calculate the accumulation of quantities, such as areas under curves or the work done by a force.
In the exercise, the work done by the force is represented as a definite integral, which calculates the total work from one point to another along a path.In mathematical terms, an integral accumulates the area under the curve of a given function from a starting point to an endpoint.
For a variable force described by the function \( F(x) = \frac{b}{x^n} \), the integral \( W = \int_{x_0}^{\infty} F(x) \, dx \) computes how much work the force does as the particle moves from point \( x_0 \) towards infinity.
When solving the integral, the solution gives insight into how much work is positive, finite, or infinite.
  • For \( n > 1 \), the result is finite because the power of \( x \) in \( x^{1-n} \) reduces the effect of the larger values as \( x \) approaches infinity.
  • For \( 0 < n < 1 \), the integral diverges and results in infinite work, illustrating how even a diminishing force can accumulate significant work over an infinite distance.
Variable force
A variable force is a force whose magnitude and possibly direction change based on position, time, or another variable.
In the context of the exercise, the force is variable as it depends on \( x \) and is expressed as \( F(x) = \frac{b}{x^n} \). This indicates that the force changes depending on the position along the x-axis.Understanding how a variable force acts is crucial because it requires careful consideration of where and how the force varies when calculating work over a distance.
Integrals are particularly useful in this situation, as they handle the accumulation of work done by such forces as they vary along the path of motion.
In our exercise, the variation in force creates different scenarios for different values of \( n \).
  • When \( n > 1 \), the force diminishes rapidly as \( x \) increases, leading to a finite amount of work as the result.
  • When \( 0 < n < 1 \), the force diminishes more slowly, causing an infinite accumulation of work as the particle travels to infinity.
This demonstrates the profound impact variable forces have on the calculations of work and energy. By understanding the influence of these forces, students can better grasp the complexities of real-world physics scenarios.

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Most popular questions from this chapter

You are asked to design spring bumpers for the walls of a parking garage. A freely rolling \(1200-\mathrm{kg}\) car moving at 0.65 \(\mathrm{m} / \mathrm{s}\) is to compress the spring no more than 0.070 \(\mathrm{m}\) before stopping. What should be the force constant of the spring? Assume that the spring has negligible mass.

Chin-Ups. While doing a chin-up, a man lifts his body 0.40 \(\mathrm{m}\) . (a) How much work must the man do per kilogram of body mass? (b) The muscles involved in doing a chin-up can generate about 70 \(\mathrm{J}\) of work per kilogram of muscle mass. If the man can just barely do a \(0.40-\mathrm{m}\) chin-up, what percentage of his body's mass do these muscles constitute? (For comparison, the total percentage of muscle in a typical \(70-k g\) man with 14\(\%\) body fat is about 43\(\%\) . (c) Repeat part (b) for the man's young son, who has arms half as long as his father's but whose muscles can also generate 70 \(\mathrm{J}\) of work per kilogram of muscle mass. (d) Adults and children have about the same percentage of muscle in their bodies. Explain why children can commonly do chin-ups more easily than their fathers.

How many joules of energy does a 100 -watt light bulb use per hour? How fast would a \(70-\mathrm{kg}\) person have to run to have that amount of kinetic energy?

A force of 160 \(\mathrm{N}\) stretches a spring 0.050 \(\mathrm{m}\) beyond its unstretched length. (a) What magnitude of force is required to stretch the spring 0.015 \(\mathrm{m}\) beyond its unstretched length? To compress the spring 0.020 \(\mathrm{m} ?\) (b) How much work must be done to stretch the spring 0.015 \(\mathrm{m}\) beyond its unstretched length? To compress the spring 0.020 \(\mathrm{m}\) from its unstretched length?

The mass of a proton is 1836 times the mass of an electron. (a) A proton is traveling at speed \(V\) . At what speed (in terms of \(V )\) would an electron have the same kinetic energy as the proton? (b) An electron has kinetic energy \(K\) . If a proton has the same speed as the electron, what is its kinetic energy (in terms of \(K ) ?\)
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