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 - Practice Exercises - Page 363: 25

Answer

$4640 J$

Work Step by Step

Integrate the integral to calculate the work done area as follows: Work done, $W=w_1+w_2= \int_{0}^{100}(100) dx+ \int_{0}^{40} (0.8) (40-x) dx$ or, $=0.8 \times [40 x-\dfrac{x^2}{2}]_0^{40}$ or, $= \dfrac{(0.8) (1600)}{2} $ or, $=640$ J Total work done: $W=4800 + 640 =4640 J$
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