Thomas' Calculus 13th Edition

Published by Pearson
ISBN 10: 0-32187-896-5
ISBN 13: 978-0-32187-896-0

Chapter 12: Vectors and the Geometry of Space - Section 12.3 - The Dot Product - Exercises 12.3 - Page 713: 44


$2,640,000\ ft\cdot lb$

Work Step by Step

The work done by a constant force ${\bf F}$ acting through a displacement ${\bf D}=\vec{PQ}$ is ${\bf W}={\bf F}\cdot{\bf D}$ --- ${\bf F}\cdot{\bf D}=|{\bf F}|\cdot|{\bf D}|\cos\alpha,\quad 1\ mi=5280\ ft.$ ${\bf W}=(1000\ lb)(5280\ ft)(\cos 60^{\circ})$ $=5,280,000\displaystyle \cdot\frac{1}{2}\ ft\cdot lb$ $=2,640,000\ ft\cdot lb$
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