Elasticity, Hooke’s Law & Young’s Modulus

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Elasticity: Hooke’s Law & Young’s Modulus

Detailed Note on Elasticity

Elasticity is the property that allows materials to return to their original shape after deformation. This lesson covers Hooke’s Law, Young’s modulus, stress, strain, and worked examples.

Hooke's Law

F = kx

Where k is the spring constant (N/m), F is force (N) and x is extension (m).

Young’s Modulus

Y = (F × L) / (A × ΔL)

Stress (σ) = F/A (Pa). Strain (ε) = ΔL / L (dimensionless). Young’s modulus is stress divided by strain.

10 JAMB Worked Examples

Example 1: Determine the Spring Constant

Question: A spring stretches 0.05 m when a force of 10 N is applied. Find k.

Solution:

k = F / x = 10 N / 0.05 m = 200 N/m

Example 2: Work Done in a Spring

Question: k = 150 N/m, x = 0.1 m. Calculate work done.

Solution:

W = ½ k x² = 0.5 × 150 × 0.1² = 0.75 J

Example 3: Find the Extension

Question: F = 25 N, k = 125 N/m. What is x?

Solution:

x = F / k = 25 / 125 = 0.2 m

Example 4: Young’s Modulus Calculation

Question: L₀ = 2 m, A = 0.005 m², ΔL = 0.002 m under F = 20,000 N. Find Y.

Solution:

ε = ΔL / L₀ = 0.002 / 2 = 0.001. σ = F / A = 20000 / 0.005 = 4,000,000 Pa. Y = σ / ε = 4.0 × 10⁹ Pa.

Example 5: Elastic Limit (Text)

Question: Linear up to 30 N at 0.15 m extension — what's the elastic limit (force)?

Solution:

The elastic limit (force) is 30 N — beyond this Hooke’s Law may not hold.

Example 6: Force–Extension Curve Analysis

Question: Identify yield and breaking points on the curve.

Solution:

Yield point = start of plastic deformation; Breaking point = point of fracture where the curve falls.

Example 7: Spring Balance Reading

Question: A balance shows extension matching 15 N — what force acts?

Solution:

The force acting is 15 N (spring balance reading gives force directly when calibrated).

Example 8: Work per Unit Volume

Question: Stress = 50,000 Pa, strain = 0.02. Compute work/volume.

Solution:

Work per unit volume = ½ × stress × strain = 0.5 × 50000 × 0.02 = 500 N/m² (J/m³).

Example 9: Stress, Strain & Y

Question: L₀=1.5 m, A=0.002 m², ΔL=0.003 m, F=1200 N. Find σ, ε, Y.

Solution:

ε=0.003/1.5=0.002. σ=1200/0.002=600,000 Pa. Y=σ/ε=600,000/0.002=3.0×10⁸ Pa.

Example 10: Identify Key Points

Question: Identify elastic limit, yield point, and breaking point from a typical curve.

Solution:

Elastic limit: end of linear region. Yield: deviation from linearity (permanent deformation). Breaking: fracture point.

30-Q JAMB CBT Quiz

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