## College Physics (4th Edition)

Published by McGraw-Hill Education

# Chapter 10 - Problems - Page 397: 11

#### Answer

The string should be stretched by $5.0~mm$

#### 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 We can find the required change in length: $\Delta L = \frac{F~L}{Y~A}$ $\Delta L = \frac{(20~N)(0.50~m)}{(2.0\times 10^9~N/m^2)(1.0\times 10^{-6}~m^2)}$ $\Delta L = 5.0\times 10^{-3}~m = 5.0~mm$ The string should be stretched by $5.0~mm$

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