A body of mass 11kg is suspended from a ceiling fan by an aluminum wire of length 2m and diameter 2mm…

Analyze the question: A wire stretched by a hanging load stores elastic energy. We need that energy.

Concept: Elastic energy = ½ × F × e, where e is the extension. Young's modulus: E = (F/A) / (e/L), so e = FL/(AE). Therefore, energy = F²L / (2AE).

Solution:

• Force: F = mg = 11 × 10 = 110 N
• Radius: r = 1 mm = 1 × 10⁻³ m
• Area: A = πr² = 3.142 × 10⁻⁶ m²
• Extension: e = FL/(AE) = (110 × 2) / (3.142 × 10⁻⁶ × 7.0 × 10¹⁰) = 220 / (2.199 × 10⁵) ≈ 1.0 × 10⁻³ m
• Energy: ½ × 110 × 1.0 × 10⁻³ = 5.5 × 10⁻² J

More precisely: e = 220 / 219,940 ≈ 1.0003 × 10⁻³ m, so energy = 55 × 1.0003 × 10⁻³ ≈ 5.5 × 10⁻² J.

5 × 10⁻² J.

Tip: Elastic energy can be written three ways: ½Fe, F²L/(2AE), or ½ × stress × strain × volume. Remember to convert diameter to radius, and mm to m. Forgetting to halve the diameter is a common mistake.