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Chapter 7: Problem 881

If the interatomic spacing in a steel wire is \(3 \AA \&\) $\mathrm{Y}_{\text {Steel }}=20 \times 10^{10}\left(\mathrm{~N} / \mathrm{m}^{2}\right) .$ The force constant \(=\ldots \ldots \ldots .\) (A) \(6 \times 10^{-2}(\mathrm{~N} / \AA)\) (B) \(6 \times 10^{-9}(\mathrm{~N} / \AA)\) (C) \(6 \times 10^{-5}(\mathrm{~N} / \AA)\) (D) \(6 \times 10^{-5}(\mathrm{~N} / \AA)\)

Short Answer

Expert verified
The force constant (k) in a steel wire can be calculated using Hooke's Law and Young's modulus(Y), along with the given interatomic spacing. After converting the units accordingly and performing the necessary calculations, the force constant is found to be \(k = 6 \times 10^{-9}\,N/\AA\). Thus, the correct answer is (B) \(6 \times 10^{-9}\,N/\AA\).

Step by step solution

01

Understand Hooke's Law

Hooke's Law states that the force acting on a spring (or similar elastic object) is directly proportional to its displacement from the equilibrium position, i.e., \(F = kx\), where F is the force, k is the force constant, and x is the displacement.
02

Calculate the strain

Strain (\(ε\)) is defined as the relative change in length due to stress, i.e., \(\epsilon = \frac{x}{L}\), where L is the initial length of the wire.
03

Use Young's modulus

Young's modulus (Y) is the ratio of stress to strain, i.e., \(Y = \frac{F}{A\epsilon}\), where A is the cross-sectional area of the wire. Rearranging this equation to solve for F gives \(F = Y \cdot A \cdot \epsilon\).
04

Calculate the force constant

Given that Hooke's Law states \(F = kx\), we can set the expressions for F from Steps 3 and 4 equal to each other: \(Y \cdot A \cdot \epsilon = kx\). We can then solve for k, the force constant, by dividing both sides by x: \(k = \frac{Y \cdot A \cdot \epsilon}{x}\).
05

Implement the given values into the formula

Now, we can plug in the provided values for Y, A, and x: \(k = \frac{(20 \times 10^{10}\,N/m^2) \cdot (3 \times 10^{-10}\,m)}{x}\). Because the force constant must have units of N/Å, we would like to express the values in compatible units. To do this, we first convert the provided units of Y into compatible units, changing \(\frac{N}{m^{2}}\) into \(\frac{N}{\AA^{2}}\), which gives us: \(Y=20\times10^{10}\times10^{-20}=2\times10^{-8} \frac{N}{\AA^{2}}\). Next, we substitute these units into the formula: \(k = \frac{(2\times10^{-8}\,N/\AA^2) \cdot (3 \times 10^{-10}\,m)}{x}\).
06

Convert the displacement x to angstroms

To finally get the force constant in the desired N/Å units, we need to express the displacement x in angstroms. As 1 Å = \(1 \times 10^{-10}\,m\), we can set \(\frac{x}{L} = \frac{x \text{ (in Å)}}{(3 \times 10^{-10}\,m)}\). Therefore, \(x_{in \AA} = \frac{x}{1 \times 10^{-10}}\), which when substituted into the formula for k, gives us: \(k = \frac{(2\times10^{-8}\,N/\AA^2) \cdot (3 \times 10^{-10}\,m)}{1\times10^{-10}\cdot x}\).
07

Calculate k

By canceling out the \(x\) value and completing the calculation, we arrive at the force constant: \(k = 6 \times 10^{-9}\,N/\AA\). Therefore, the correct answer is (B) \(6 \times 10^{-9}\,N/\AA\).

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