Physics for Scientists and Engineers: A Strategic Approach with Modern Physics (4th Edition)

Published by Pearson
ISBN 10: 0133942651
ISBN 13: 978-0-13394-265-1

Chapter 9 - Work and Kinetic Energy - Exercises and Problems - Page 229: 50


$v = 1.9~m/s$

Work Step by Step

The work done on the fish by the tension will be equal to the sum of the kinetic energy and the potential energy. Therefore; $KE+PE = W_T$ $\frac{1}{2}mv^2+mgh = T~d$ $v^2 = \frac{2(T~d-mgh)}{m}$ $v = \sqrt{\frac{2~(T~d-mgh)}{m}}$ $v = \sqrt{\frac{(2)~[(16~N)(2.0~m)-(1.5~kg)(9.80~m/s^2)(2.0~m)]}{1.5~kg}}$ $v = 1.9~m/s$
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