Calculus (3rd Edition)

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

Chapter 6 - Applications of the Integral - 6.5 Work and Energy - Exercises - Page 316: 2


$$64 \ lb\cdot ft.$$

Work Step by Step

The force needed is $4 \ lb$. Hence, the work is given by ($W=Fd$): $$(4 \ lb)(16 \ ft)=64 \ lb\cdot ft.$$
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