Calculus 10th Edition

Published by Brooks Cole
ISBN 10: 1-28505-709-0
ISBN 13: 978-1-28505-709-5

Chapter 7 - Applications of Integration - 7.5 Exercises - Page 483: 4

Answer

$$W = 47,520,000{\text{ ft - lb}}$$

Work Step by Step

$$\eqalign{ & {\text{From the information we have:}} \cr & F = {\text{9tons }}\left( {{\text{force}}} \right) \cr & D = \frac{1}{2}{\text{mile }}\left( {{\text{distance}}} \right) \cr & {\text{Unit conversions}} \cr & {\text{9ton}}\left( {\frac{{2000{\text{lb}}}}{{1{\text{ton}}}}} \right) = 18,000{\text{lb}} \cr & \frac{1}{2}{\text{mile}}\left( {\frac{{5280{\text{ft}}}}{{1{\text{mile}}}}} \right) = 2,640{\text{ft}} \cr & {\text{Using the Definition of Work Done by a Constant Force}} \cr & W = FD{\text{, Work}} = \left( {{\text{force}}} \right)\left( {{\text{distance}}} \right) \cr & {\text{Substituting the known values}} \cr & W = \left( {18,000{\text{lb}}} \right)\left( {2,640{\text{ft}}} \right) \cr & W = 47,520,000{\text{ ft - lb}} \cr} $$
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