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: 27


$3920 \ J$

Work Step by Step

The force of one layer is equal to: $\ Force= \ Mass \times \ gravity = 78.4 \Delta y \ N$ The work done on the $i$th segment is the sum of the $W_i$; that is, $W=\Sigma_{j=1}^N W_i \approx \Sigma_{j=1}^N 78.4 y_j \Delta y$ Therefore, the work done when $N \to \infty $ can be computed as: $ W=\int_{0}^{10} 78.4 \ y \ dy \\=39.2 [y^2]_{0}^{10} \\ =3920 \ J$
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