University Calculus: Early Transcendentals (3rd Edition)

Published by Pearson
ISBN 10: 0321999584
ISBN 13: 978-0-32199-958-0

Chapter 6 - Practice Exercises - Page 391: 26


$$22,800,00 \space ft- lb$$

Work Step by Step

Work done, $W=\int_c^d F(x) dx$ We have : $F(x)= 8 \times 800 [\dfrac{(2) \times 4750-x}{(2) \times (4750)}] $ or, $$F(x)=6400 (1- \dfrac{x}{9500} ) lb$$ Now, $$W= (6400) \times \int_{0}^{4750} (1-\dfrac{x}{9500}) dx \\=6400 \times [4750-\dfrac{(4750)^2}{4 \times 4750 }] \\=22,800,00 \space ft- lb$$
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