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 318: 33

Answer

$374.85 \ J$

Work Step by Step

We calculate the work done to lift the 3-m chain by integrating as follows: $ Work \ done =\int_{0}^{3} (9.8) (k(x)) \ x \ dx \\=9.8 \int_0^3 (x^3-3x^2+10 x) \ dx \\=9.8 \times [\dfrac{x^4}{4}-x^3+5x^2]_{0}^{3}\\=9.8 \times [\dfrac{(3)^4}{4}-(3)^3+5(3)^2] \\=374.85 \ J$
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