Skip to content
MarkSchemeMarkScheme
Sign up — no card

9702 · 6.2

Elastic and plastic behaviour — practice questions

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

Worked example 1

A spring has a spring constant of 250 N m⁻¹. Calculate the elastic potential energy stored in the spring when it is extended by 5.0 cm.

Show solution outline
  1. Convert extension to metres: ΔL=5.0 cm=0.050 m\Delta L = 5.0\text{ cm} = 0.050\text{ m}.
  2. Use the formula for EPE involving kk and ΔL\Delta L: EPE=12k(ΔL)2EPE = \frac{1}{2}k(\Delta L)^2
  3. Substitute the values: EPE=12×250 N m−1×(0.050 m)2EPE = \frac{1}{2} \times 250\text{ N m}^{-1} \times (0.050\text{ m})^2
  4. Calculate the result: EPE=12×250×0.0025=0.3125 JEPE = \frac{1}{2} \times 250 \times 0.0025 = 0.3125\text{ J} Therefore, the elastic potential energy stored is 0.31 J (to 2 s.f.).

Worked example 2

A steel wire of original length 2.0 m and diameter 0.50 mm is stretched by a force of 40 N. The extension produced is 2.5 mm. Calculate: (a) the stress in the wire, (b) the strain of the wire, and (c) the Young Modulus of steel.

Show solution outline
  1. List knowns and convert to SI units: Force, F=40F = 40 N Original length, L0=2.0L_0 = 2.0 m Extension, ΔL=2.5 mm=2.5×10−3\Delta L = 2.5 \text{ mm} = 2.5 \times 10^{-3} m Diameter, d=0.50 mm=0.50×10−3d = 0.50 \text{ mm} = 0.50 \times 10^{-3} m Radius, r=d/2=0.25×10−3r = d/2 = 0.25 \times 10^{-3} m
  2. Calculate cross-sectional area (A): A=πr2=π(0.25×10−3)2=1.963×10−7 m2A = \pi r^2 = \pi (0.25 \times 10^{-3})^2 = 1.963 \times 10^{-7} \text{ m}^2
  3. (a) Calculate Stress (σ\sigma): σ=FA=40 N1.963×10−7 m2=2.037×108 Pa\sigma = \frac{F}{A} = \frac{40 \text{ N}}{1.963 \times 10^{-7} \text{ m}^2} = 2.037 \times 10^8 \text{ Pa} Stress ≈2.0×108\approx 2.0 \times 10^8 Pa (or 200 MPa)
  4. (b) Calculate Strain (ϵ\epsilon): ϵ=ΔLL0=2.5×10−3 m2.0 m=1.25×10−3\epsilon = \frac{\Delta L}{L_0} = \frac{2.5 \times 10^{-3} \text{ m}}{2.0 \text{ m}} = 1.25 \times 10^{-3} Strain is dimensionless.
  5. (c) Calculate Young Modulus (E): E=σϵ=2.037×108 Pa1.25×10−3=1.6296×1011 PaE = \frac{\sigma}{\epsilon} = \frac{2.037 \times 10^8 \text{ Pa}}{1.25 \times 10^{-3}} = 1.6296 \times 10^{11} \text{ Pa} Young Modulus ≈1.6×1011\approx 1.6 \times 10^{11} Pa (or 160 GPa)
Mark a 9702 question on this topicFull lessonOpen in course studio