Thomas' Calculus 13th Edition

Published by Pearson
ISBN 10: 0-32187-896-5
ISBN 13: 978-0-32187-896-0

Chapter 6: Applications of Definite Integrals - Section 6.5 - Work and Fluid Forces - Exercises 6.5 - Page 347: 2


a) $200$ \lb/in b) $400 lb- in. \approx 33.3 ft -lb$ c) 8 in.

Work Step by Step

a. Hooke's Law states that $F=k x$ and $k=\dfrac{F}{x}=\dfrac{800}{4}= 200$ \lb/in b. Need to integrate $k =200$ with limit $0$ to $2$ Use formula: $\int x^n=\dfrac{x^{n+1}}{n+1}+C$ Thus, $\int_0^2 kx dx= 200[ \dfrac{x^2}{2}]_0^2$ so, $100(4-0)=400 lb- in. \approx 33.3 ft -lb$ c. Hooke's Law states that $F=k x$ Given: $ F= 1600 N$ Then, $1600=(200) (x)$ and $x=\dfrac{1600}{200}=8$ inch
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