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 10 - Interactions and Potential Energy - Exercises and Problems: 20

Answer

The plane's landing speed was 60 m/s

Work Step by Step

The plane's kinetic energy when it lands will be equal to the energy $U_s$ stored in the spring. We can find the landing speed of the plane. Therefore; $KE = U_s$ $\frac{1}{2}mv^2=\frac{1}{2}kx^2$ $v^2=\frac{kx^2}{m}$ $v=\sqrt{\frac{kx^2}{m}}$ $v=\sqrt{\frac{(60,000~N/m)(30~m)^2}{15,000~kg}}$ $v = 60~m/s$ The plane's landing speed was 60 m/s.
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