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Ch. 6 - Applications of Definite Integrals
Hass - Thomas' Calculus 15th Edition
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Tutti i libri di testoHass 15th EditionCh. 6 - Applications of Definite IntegralsProblema 6.PE.35a
Capitolo 6, Problema 6.PE.35a

Work


Assume that a spring does not follow Hooke’s Law. Instead, the force required to stretch the spring x ft from its natural length is ƒ(𝓍) = 10𝓍³/² lb . How much work does it take to
a. stretch the spring 4 ft from its natural length?

Guida verificata passo dopo passo
1
Identify the force function given as a function of displacement \(x\): \(f(x) = 10x^{3/2}\) pounds, where \(x\) is the distance stretched from the natural length in feet.
Recall that the work done to stretch the spring from \(x = a\) to \(x = b\) is given by the integral of the force over that interval: \(W = \int_a^b f(x) \, dx\).
Since the spring is stretched from its natural length (which corresponds to \(x = 0\)) to 4 feet, set the limits of integration as \(0\) to \(4\).
Set up the integral for work: \(W = \int_0^4 10x^{3/2} \, dx\).
To solve the integral, use the power rule for integration: \(\int x^n \, dx = \frac{x^{n+1}}{n+1} + C\). Apply this to \(x^{3/2}\) and then multiply by 10.

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Concetti chiave

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Work Done by a Variable Force

Work done by a force that varies with position is calculated by integrating the force function over the displacement interval. Unlike constant forces, variable forces require setting up an integral to sum the infinitesimal amounts of work done over small displacements.
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Percorso guidato
05:40
Work Done On A Spring (Hooke's Law)

Force as a Function of Displacement

In this problem, the force is given as a function of displacement, ƒ(x) = 10x^(3/2). Understanding how force depends on position is crucial to correctly formulating the integral for work, especially when the force does not follow Hooke’s Law.
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Percorso guidato
10:17
Using The Velocity Function

Definite Integration for Physical Quantities

Definite integrals are used to compute total quantities like work over a specific interval. Here, integrating the force function from 0 to 4 ft will yield the total work done in stretching the spring from its natural length to 4 ft.
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Percorso guidato
05:43
Definition of the Definite Integral
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