Essential University Physics: Volume 1 (3rd Edition)

Published by Pearson
ISBN 10: 0321993721
ISBN 13: 978-0-32199-372-4

Chapter 14 - Exercises and Problems: 29

Answer

Please see the work below.

Work Step by Step

We know that $v=\sqrt{\frac{F}{\mu}}$ We plug in the known values to obtain: $v=\sqrt{\frac{550}{0.280}}=44.32\frac{m}{s}$ Now we can find the average power as $P=(\frac{1}{2})\mu \omega^2A^2v$ We plug in the known value to obtain: $P=(\frac{1}{2})(0.280)(2\times 3.1416\times 3.3)^2(0.061)^2(44.32)$ $P=9.9W$
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