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: 30


$1960 \ J$

Work Step by Step

The work required to lift the segment of the chain to the top of the building can be computed as: $W_{i} \approx (4 \Delta y) (9.8 y_i) \ J$ Therefore, the total work done when $\Delta y \to 0$ can be computed as: $ W=\int_{2}^{10} 39.2 \ y \ dy \\=39.2 [\dfrac{y^2}{2}]_{2}^{10}\\=19.6 9 \times [y^2]_{2}^{10} \\= 1960 \ J$
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