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 - Conceptual Questions: 12

Answer

A force of 4000 N is needed to stretch the wire by 1 mm

Work Step by Step

We can use Young's modulus to solve this question. $Y = \frac{F/A}{\Delta~L/L}$ We can find an expression for the force required to stretch a wire. $Y = \frac{F/A}{\Delta~L/L}$ $F = \frac{Y~A~\Delta L}{L}$ $F = \frac{Y~\pi~R^2~\Delta L}{L}$ We can write an expression for $F_1$ $F_1 = \frac{Y~\pi~R_1^2~\Delta L}{L_1}$ We can write an expression for $F_2$ $F_2 = \frac{Y~\pi~(2R_1)^2~\Delta L}{2L_1}$ $F_2 = 2\times\frac{Y~\pi~R_1^2~\Delta L}{L_1}$ $F_2 = 2~F_1$ $F_2 = (2)(2000~N)$ $F_2 = 4000~N$ A force of 4000 N is needed to stretch the wire by 1 mm
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