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Ch 09: Work and Kinetic Energy
Knight Calc - Physics for Scientists and Engineers 5th Edition
Knight Calc5th EditionPhysics for Scientists and EngineersISBN: 9780137344796Not the one you use?Change textbook
Chapter 9, Problem 23

A particle moving on the x-axis experiences a force given by Fx = qx², where q is a constant. How much work is done on the particle as it moves from x = 0 to x = d?

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The work done on a particle by a force is calculated using the work integral: \( W = \int F_x \, dx \). Here, \( F_x = qx^2 \), and the limits of integration are from \( x = 0 \) to \( x = d \).
Substitute \( F_x = qx^2 \) into the integral: \( W = \int_{0}^{d} qx^2 \, dx \).
Since \( q \) is a constant, it can be factored out of the integral: \( W = q \int_{0}^{d} x^2 \, dx \).
Evaluate the integral of \( x^2 \): \( \int x^2 \, dx = \frac{x^3}{3} \). Apply this result to the limits of integration: \( W = q \left[ \frac{x^3}{3} \right]_{0}^{d} \).
Substitute the limits \( x = d \) and \( x = 0 \) into the expression: \( W = q \left( \frac{d^3}{3} - \frac{0^3}{3} \right) \). Simplify the result to express the work done in terms of \( q \) and \( d \).

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Key Concepts

Here are the essential concepts you must grasp in order to answer the question correctly.

Work Done by a Force

Work is defined as the integral of force over the distance moved in the direction of the force. Mathematically, it is expressed as W = ∫ F dx. In this case, the force Fx = qx² varies with position, so the work done as the particle moves from x=0 to x=d requires evaluating the integral of this force function over that interval.
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Integration in Physics

Integration is a fundamental mathematical tool used in physics to calculate quantities that accumulate over a continuous range. In the context of work, it allows us to sum up the infinitesimal contributions of force over a distance. For the given force function, integrating Fx = qx² from 0 to d will yield the total work done on the particle.
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Variable Force

A variable force is one that changes in magnitude and/or direction over time or space. In this problem, the force acting on the particle depends on its position (x) and is described by the equation Fx = qx². Understanding how to handle variable forces is crucial for accurately calculating work, as it requires integration rather than simple multiplication of force and distance.
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