Problem 9 - College Physics

Question

A \(200-\mathrm{kg}\) load is hung on a wire of length \(4.00 \mathrm{~m}\), cross-sectional area \(0.200 \times 10^{-4} \mathrm{~m}^{2}\), and Young's modulus \(8.00 \times 10^{10} \mathrm{~N} / \mathrm{m}^{2}\). What is its increase in length?

Step-by-Step Solution

Answer

\(\Delta L = 0.0098 \: m\) or \(9.8 \:mm\).

Step 1: Understand the Problem

The task is to find the increase in length - often called elongation - of a wire when a certain load is applied. The properties of the wire, including its original length, cross-sectional area, and elasticity (Young's modulus) are given, as well as the weight that is hung on it. The formula that connects all these values is that for Young's modulus, \(Y = \frac{F}{A}/\frac{\Delta L}{L}\), where \(F\) is the force on the wire, \(A\) is the cross-sectional area of the wire, \(\Delta L\) is the change in length (elongation), and \(L\) is the original length. Rearrange the formula, isolating \(\Delta L\), to solve for elongation.

Step 2: Substitute the values into the formula

After rearranging the formula for Young's modulus, you get \(\Delta L = \frac{FL}{YA}\). Substitute the given values into the formula: \( F = 200 kg \times 9.81 m/s² = 1962 N\) (force equals mass times gravity), \( L = 4 m \), \( Y = 8.00 \times 10^{10} N/m² \) and \( A = 0.200 \times 10^{-4} m² \).

Step 3: Calculation

Now perform the multiplication and division operations as indicated by the formula. This operation should yield the increase in length of the wire, \(\Delta L\), when a 200-kg weight is hung from it.

Key Concepts

Elasticity

Elasticity refers to a material’s ability to return to its original shape after being stretched or compressed. It's like a rubber band snapping back after you pull on it. In physics, elasticity is quantified using Young's modulus, a constant that measures how much a material will deform under stress. A high Young's modulus means the material is rigid, while a low value indicates flexibility. Consider Young's modulus as a stiffness indicator that provides insight into how different materials behave when forces are applied.

Stress and Strain

Stress and strain are central to understanding how materials react to forces. Stress is the force exerted on a material per unit area, calculated as \( \text{Stress} = \frac{F}{A} \), where \( F \) is the force on the object and \( A \) is the cross-sectional area. Strain, on the other hand, is the deformation or elongation of the material, determined by \( \text{Strain} = \frac{\Delta L}{L} \), where \( \Delta L \) is the change in length, and \( L \) is the original length.

These two concepts help predict how much a material will stretch or compress when a force is applied. Using Young's modulus, we can link stress and strain with the equation: \( Y = \frac{\text{Stress}}{\text{Strain}} \). This relationship is key in solving physics problems related to material deformation.

Physics Problem Solving

Solving physics problems requires a clear understanding of the concepts and how to apply given formulas. Let's explore a typical approach using the example of calculating the elongation of a wire. Start by understanding what is known:

• Original length of the wire \( (L) \)
• Cross-sectional area \( (A) \)
• Applied force \( (F) \)
• Young's modulus \( (Y) \)

Next, rearrange the Young's modulus formula to solve for the unknown, typically the change in length \( (\Delta L) \): \( \Delta L = \frac{FL}{YA} \). Substitute the known values into the equation and perform the calculations step-by-step.

Check your work to ensure logical conclusions, validating that the numerical results make sense given the physical context. This practical problem-solving strategy keeps complicated tasks manageable and clear.