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: 5

Answer

a) $k=7238$ lb/in. b) $W=2714$ in/lb

Work Step by Step

a. Hooke's Law states that $F=k x$ or,$k=\dfrac{F}{x}=\dfrac{21714}{3}=7238$ lb/in. $\implies F=7238 x$ b. i) Use formula: $\int x^n=\dfrac{x^{n+1}}{n+1}+C$ This implies that $W=\int_0^{0.5} 7238 x dx=7238[ \dfrac{x^2}{2}]_0^{0.5}$ or, $3619[ (0.5)^2-0] =904.75 N-m$ ii) Use formula: $\int x^n=\dfrac{x^{n+1}}{n+1}+C$ This implies that $W=\int_{0.5}^1 7238 x dx=7238[ \dfrac{x^2}{2}]_{0.5}^1$ so, $3619[1- (0.5)^2] =2714 in/lb$
Update this answer!

You can help us out by revising, improving and updating this answer.

Update this answer

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.