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: 29


$529.2 \ J$

Work Step by Step

The work required to lift the segment of the chain to the top of the building can be computed as: $W_{i}=\Sigma_{i=1}^N W_i \approx (3 \Delta y) (9.8) y_i \ ft-lb$ Therefore, the total work done when $\Delta y \to 0$ can be computed as: $ W=\int_{0}^{6} 29.4 \ y \ dy \\=14.7[y^2]_{0}^{6}\\=14.7 \times 36 \\=529.2 \ J$
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