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 317: 16

Answer

$141.26 \ k \ J$

Work Step by Step

The force of one layer is equal to: $\ Force= \ Mass \times \ gravity = 39.24 (1000-100 y) \Delta y \ N$ Therefore, the work done against can be computed as: $ W=39.24 \int_{0}^{3} (1000-100 y) \times y \ d y\\= 39.24 \int_{0}^{3} (1000y-100y^2) \ d y\\ = 39.24 [\dfrac{1000y^2}{2}-\dfrac{100 y^3}{3} ]_0^{3} $ By using a calculator, the required result is: $W =141.26 \ k \ J$
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