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: 2

Answer

$$W = 15,000{\text{ ft - lb}}$$

Work Step by Step

$$\eqalign{ & {\text{From the information we have:}} \cr & F = {\text{2500 - pounds }}\left( {{\text{force}}} \right) \cr & D = 6{\text{ feet }}\left( {{\text{distance}}} \right) \cr & {\text{Using the Definition of Work Done by a Constant Force}} \cr & W = FD{\text{, Work}} = \left( {{\text{force}}} \right)\left( {{\text{distance}}} \right) \cr & {\text{Substituting the known values}} \cr & W = \left( {{\text{2500}}} \right)\left( {\text{6}} \right) \cr & W = 15,000{\text{ ft - lb}} \cr} $$
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