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9702 · 6.1

Stress and strain — practice questions

Practice and worked examples for 9702 Stress and strain. Short previews only — attempt the full question in MarkScheme against the official scheme.

Worked example 1

A copper wire has an original length of 2.0 m and a cross-sectional area of 1.5 x 10−7^{-7} m2^2. It is stretched by a tensile force of 45 N. Given that the Young Modulus for copper is 1.1 x 1011^{11} Pa, calculate the extension of the wire.

Show solution outline
  1. Recall the formula for Young Modulus: E=FL0AΔLE = \frac{F L_0}{A \Delta L}.
  2. Rearrange the formula to solve for extension ΔL\Delta L: ΔL=FL0AE\Delta L = \frac{F L_0}{A E}.
  3. Substitute the given values into the rearranged formula:
    • Force (FF) = 45 N
    • Original Length (L0L_0) = 2.0 m
    • Cross-sectional Area (AA) = 1.5 x 10−7^{-7} m2^2
    • Young Modulus (EE) = 1.1 x 1011^{11} Pa
  4. Perform the calculation: ΔL=(45 N)(2.0 m)(1.5×10−7 m2)(1.1×1011 Pa)\Delta L = \frac{(45 \text{ N})(2.0 \text{ m})}{(1.5 \times 10^{-7} \text{ m}^2)(1.1 \times 10^{11} \text{ Pa})} ΔL=9016500≈0.00545 m\Delta L = \frac{90}{16500} \approx 0.00545 \text{ m}
  5. The extension of the wire is approximately 5.5 x 10−3^{-3} m (or 5.5 mm).

Worked example 2

A steel rod of length 2.5 m and cross-sectional area 8.0 x 10⁻⁵ m² is subjected to a tensile force that causes it to extend by 1.2 mm. The Young Modulus of steel is 2.0 x 10¹¹ Pa. Calculate the elastic potential energy (EPE) stored in the rod.

Show solution outline
  1. Identify the goal: Calculate Elastic Potential Energy (EPE). The formula is EPE = ½ * F * ΔL.
  2. Identify the missing variable: The tensile force (F) is not given. We must calculate it first.
  3. Use the Young Modulus formula to find F: E = (F * L₀) / (A * ΔL). Rearrange for F: F = (E * A * ΔL) / L₀.
  4. Ensure all units are SI. Convert extension ΔL from mm to m: ΔL = 1.2 mm = 1.2 x 10⁻³ m.
  5. Substitute the values to calculate F:
    • E = 2.0 x 10¹¹ Pa
    • A = 8.0 x 10⁻⁵ m²
    • ΔL = 1.2 x 10⁻³ m
    • L₀ = 2.5 m F = ( (2.0 x 10¹¹) * (8.0 x 10⁻⁵) * (1.2 x 10⁻³) ) / 2.5 F = (19.2 x 10³) / 2.5 = 7680 N
  6. Now calculate the EPE using the calculated force: EPE = ½ * F * ΔL EPE = 0.5 * 7680 N * (1.2 x 10⁻³ m) EPE = 4.608 J
  7. The elastic potential energy stored in the rod is 4.6 J (to 2 significant figures).
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