University Calculus: Early Transcendentals (3rd Edition)

Published by Pearson
ISBN 10: 0321999584
ISBN 13: 978-0-32199-958-0

Chapter 6 - Section 6.5 - Work - Exercises - Page 380: 4


$1125$ N-m

Work Step by Step

We know that Hooke's Law states that $F=k x$ Then, $k=\dfrac{F}{x}=\dfrac{90}{01}=90$ To calculate the work done, the limits for $x$ will be from $0$ to $5$ such that: $W=\int_0^{5} 90 x dx= 90[ \dfrac{x^2}{2}]_0^{5}$ or, $W=45[ (5)^2-0] =1125$ N-m
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