Essential University Physics: Volume 1 (4th Edition) Clone

Published by Pearson
ISBN 10: 0-134-98855-8
ISBN 13: 978-0-13498-855-9

Chapter 7 - Exercises and Problems - Page 131: 46

Answer

$5.49 \ km/h$

Work Step by Step

We use conservation of energy to find: $\frac{1}{2}mv^2 = \frac{1}{2}kx^2 $ $v=\sqrt{\frac{kx^2}{m}}=\sqrt{\frac{(1.3\times10^6N)(.05m)^2}{1400}}=1.52 \ m/s = 5.49 \ km/h$
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