/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 35 The force on a particle is direc... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

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 is \( x_0 F_0 \).

Step by step solution

01

Understand the Force Function

The force on the particle is given by \( F = F_0 \left( \frac{x}{x_0} - 1 \right) \). This is a linear function in terms of \( x \), and it's important to note its nature: it varies about the point \( x_0 \). At \( x = x_0 \), the force \( F = 0 \).
02

Plot the Force Function

To plot the force as a function of \( x \), evaluate \( F(x) \) at important points. For example:- At \( x = 0 \), \( F = -F_0 \).- At \( x = x_0 \), \( F = 0 \).- At \( x = 2x_0 \), \( F = F_0 \).Plot these points and connect them linearly to get the graph of \( F(x) \) from \( x = 0 \) to \( x = 2x_0 \).
03

Measure Work from the Graph

The work done by a force is the area under the \( F(x) \) curve between the given limits. On the plot from Step 2, this area forms a trapezoid with bases at heights \( -F_0 \) and \( F_0 \), and a total length from \( 0 \) to \( 2x_0 \). The area of this trapezoid is given by:\[W = \frac{1}{2} \times \text{base}_1 \times \text{base}_2 \times \text{height}\]Substituting, \(W = \frac{1}{2} \times (0 + 2x_0) \times (-F_0 + F_0)= \frac{1}{2} \times 2x_0 \times 0 + \frac{1}{2} \times 2x_0 \times 2F_0= x_0 \times F_0 \).Thus, the work done is \( x_0 F_0 \) from the graph.
04

Set Up the Integral for Work

Work done by a varying force is found by integrating the force over the distance. So, set up the integral:\[W = \int_{0}^{2x_0} F(x) \, dx = \int_{0}^{2x_0} F_0 \left( \frac{x}{x_0} - 1 \right) \, dx\]
05

Integrate the Force Function

Evaluate the integral:\[W = F_0 \left( \int_{0}^{2x_0} \frac{x}{x_0} \, dx - \int_{0}^{2x_0} 1 \, dx \right)\]- For the first integral, \( \int \frac{x}{x_0} \, dx = \frac{1}{2x_0}x^2 \)- For the second integral, \( \int 1 \, dx = x \)So,\[W = F_0 \left( \left[ \frac{1}{2x_0}x^2 \right]_0^{2x_0} - \left[x\right]_0^{2x_0} \right)= F_0 \left( \left[ 2x_0 \right] - \left[ \frac{1}{2}(2x_0)^2 - 0 \right] \right)= F_0 \left( 2x_0 - 2x_0 \right)= x_0 F_0\]Thus, the work done, using integration, is \( x_0 F_0 \).

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91影视!

Key Concepts

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

Force Function
A force function describes how the force applied to an object varies with a particular variable, like position. In our case, the force function is given by \( F = F_0 \left( \frac{x}{x_0} - 1 \right) \). Here, \( F_0 \) represents a constant force, while \( x_0 \) is a reference point along the x-axis. The function is linear, meaning it changes in a straight line as the position \( x \) changes. This is a typical way to quantify forces that aren't constant over a range. You can think of a force function as a mathematical way to capture how force behaves over distance. Understanding this relationship is crucial because the force may change during different parts of the motion.
Integration
Integration is a fundamental tool in calculus used to calculate areas under curves, among other things. When dealing with changing forces like in our problem, integration helps us find the total work done. Work is simply the integral of force over the distance traveled. In the equation \( W = \int_{0}^{2x_0} F(x) \, dx \), we're essentially piecing together small amounts of work done over minuscule slices of the motion to find the total work from start to finish. This process of summing infinitesimally small elements is what makes integration so powerful, eliminating the need to measure work in bits and pieces manually.
Area Under Curve
The area under a curve on a graph of force vs. position represents work done by the force over that range of motion. In visual terms, if you plot \( F(x) \) over the interval from \( x=0 \) to \( x=2x_0 \), the shape area between the curve and the x-axis signifies the total work. This 'area method' turns a calculus problem into a geometry problem, using shapes like trapezoids and rectangles whose areas are easy to calculate. For instance, the force curve creates a trapezoidal shape in this task, with the height determined by force values and "bases" by the range on the x-axis. Calculating this area helps us determine the whole work done without integrating manually.
Linear Force Function
A linear force function is one where the relationship between force and position is direct and proportional. In simple terms, if you graph the force function, the line will be straight. This method is linear because the force changes at a constant rate as \( x \) changes, which is reflected in the formula \( F = F_0 \left( \frac{x}{x_0} - 1 \right) \). Here, at \( x = x_0 \), the force becomes zero. From there, the force increases linearly as \( x \) moves further. Understanding linear force functions helps explain how force varies consistently with position, making it easier to predict behavior across an interval.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

A force \(\vec{F}=(4.0 \mathrm{~N}) \hat{\mathrm{i}}+c \hat{\mathrm{j}}\) acts on a particle as the particle goes through displacement \(\vec{d}=(3.0 \mathrm{~m}) \hat{\mathrm{i}}-(2.0 \mathrm{~m}) \hat{\mathrm{j}} .\) (Other forces also act on the particle.) What is \(c\) if the work done on the particle by force \(\vec{F}\) is (a) 0, (b) \(17 \mathrm{~J}\), and (c) \(-18 \mathrm{~J}\) ?

A cave rescue team lifts an injured spelunker directly upward and out of a sinkhole by means of a motor-driven cable. The lift is performed in three stages, each requiring a vertical distance of \(10.0\) m: (a) the initially stationary spelunker is accelerated to a speed of \(5.00 \mathrm{~m} / \mathrm{s} ;(\mathrm{b})\) he is then lifted at the constant speed of \(5.00 \mathrm{~m} / \mathrm{s} ;\) (c) finally he is decelerated to zero speed. How much work is done on the \(80.0 \mathrm{~kg}\) rescuee by the force lifting him during each stage?

The only force acting on a \(2.0 \mathrm{~kg}\) canister that is moving in an \(x y\) plane has a magnitude of \(5.0 \mathrm{~N}\). The canister initially has a velocity of \(4.0 \mathrm{~m} / \mathrm{s}\) in the positive \(x\) direction and some time later has a velocity of \(6.0 \mathrm{~m} / \mathrm{s}\) in the positive \(y\) direction. How much work is done on the canister by the \(5.0 \mathrm{~N}\) force during this time?

A particle moves along a straight path through displacement \(\vec{d}=(8 \mathrm{~m}) \hat{\mathrm{i}}+c \hat{\mathrm{j}}\) while force \(\vec{F}=(2 \mathrm{~N}) \hat{\mathrm{i}}-(4 \mathrm{~N}) \hat{\mathrm{j}}\) acts on it. (Other forces also act on the particle.) What is the value of \(c\) if the work done by \(\vec{F}\) on the particle is (a) zero, (b) positive, and (c) negative?

(a) At a certain instant, a particle-like object is acted on by a force \(\vec{F}=(4.0 \mathrm{~N}) \hat{\mathrm{i}}-(2.0 \mathrm{~N}) \hat{\mathrm{j}}+(9.0 \mathrm{~N}) \hat{\mathrm{k}}\) while the object's veloc- ity is \(\vec{v}=-(2.0 \mathrm{~m} / \mathrm{s}) \hat{\mathrm{i}}+(4.0 \mathrm{~m} / \mathrm{s}) \hat{\mathrm{k}}\). What is the instantaneous rate at which the force does work on the object? (b) At some other time, the velocity consists of only a \(y\) component. If the force is unchanged and the instantaneous power is \(-12 \mathrm{~W}\), what is the velocity of the object?
See all solutions

Recommended explanations on Physics Textbooks

Translational Dynamics

Read Explanation

Particle Model of Matter

Read Explanation

Energy Physics

Read Explanation

Solid State Physics

Read Explanation

Electrostatics

Read Explanation

Geometrical and Physical Optics

Read Explanation
View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.