## Essential University Physics: Volume 1 (3rd Edition)

Published by Pearson

# Chapter 7 - Exercises and Problems - Page 126: 44

#### 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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