College Physics (4th Edition)

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

Chapter 10 - Problems - Page 398: 21


The stress at the breaking point is $4.0\times 10^8~N/m^2$

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 Note that $\frac{F}{A}$ is the stress and $\frac{\Delta L}{L}$ is the strain. We can find the stress $\frac{F}{A}$: $\frac{F}{A} = Y~\frac{\Delta L}{L}$ $\frac{F}{A} = (2.0\times 10^{11}~N/m^2)~(\frac{0.20}{100})$ $\frac{F}{A} = 4.0\times 10^8~N/m^2$ The stress at the breaking point is $4.0\times 10^8~N/m^2$.
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