Calculus (3rd Edition)

Published by W. H. Freeman
ISBN 10: 1464125260
ISBN 13: 978-1-46412-526-3

Chapter 6 - Applications of the Integral - Chapter Review Exercises - Page 320: 54

Answer

$98,000 \ J$

Work Step by Step

The force required to lift the container is equal to: $ (9.8) [20+(100-0.4 y)] =(1176 - 3.92 y ) \ N$ Therefore, the work done to lift the container can be computed as: $ W= \int_{0}^{100} (1176-3.92 y) \ d y\\= (1176 y-1.96 y^2)_{0}^{100} \\=(1176)(100) -(1.96)(10000) \\= 98,000 \ J$
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