College Physics (4th Edition)

Published by McGraw-Hill Education
ISBN 10: 0073512141
ISBN 13: 978-0-07351-214-3

Chapter 10 - Problems - Page 398: 16


The ACL must be stretched by $3.2~mm$ to tear it.

Work Step by Step

$Y = \frac{F/A}{\Delta L/L}$ $Y$ is Young's modulus $F$ is the force $A$ is the cross-sectional area $\Delta L$ is the change in length $L$ is the original length In this situation, we can let $\frac{F}{A} = 1.9\times 10^8~Pa$ We can find the change in length: $\Delta L = (\frac{F}{A})~(\frac{L}{Y})$ $\Delta L = (1.9\times 10^8~Pa)~(\frac{1.0\times 10^{-2}~m}{6.0\times 10^8~Pa})$ $\Delta L = 3.2\times 10^{-3}~m = 3.2~mm$ The ACL must be stretched by $3.2~mm$ to tear it.
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