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: 32



Work Step by Step

The work required to lift the segment of the cable to the top of the can be computed as: $W_{i} \approx (15 \Delta y) (9.8 y_i) \ J$ Therefore, the total work done on the cable when $\Delta y \to 0$ can be computed as: $ W=\int_{0}^{12} 147 \ y \ dy \\=147 [\dfrac{y^2}{2}]_{0}^{12}\\=73.5 \times [y^2]_{0}^{12} \\=73.5 \times (12)^2 \\= 10584 \ J$ In addition, lifting the 500kg ball 12 meters requires the following amount of work: $W=mgh=500kg*12~m*9.8m/s^2=58800~J$ Thus, the total work required is: $58800~J+10584~J=69384~J$
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