Physics for Scientists and Engineers: A Strategic Approach with Modern Physics (3rd Edition)

Published by Pearson
ISBN 10: 0321740904
ISBN 13: 978-0-32174-090-8

Chapter 15 - Fluids and Elasticity - Exercises and Problems - Page 436: 26


The string is stretched by 1.0 cm

Work Step by Step

We can use Young's modulus to solve this question: $Y = \frac{F/A}{\Delta~L/L} = \frac{F~L}{A~\Delta L}$ For steel, $Y = 200\times 10^9~N/m^2$ We can find the the distance $\Delta L$ that the string is stretched. $Y = \frac{F~L}{A~\Delta L}$ $\Delta L = \frac{F~L}{A~Y}$ $\Delta L = \frac{(2000~N)(0.80~m)}{(\pi)(5.0\times 10^{-4}~m)^2~(200\times 10^9~N/m^2)}$ $\Delta L = 0.010~m = 1.0~cm$ We find that the string is stretched by 1.0 cm.
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