A close-coiled helical spring absorbs 30 N·mm of energy while extending by 5 mm. Its stiffness will be:
Try it first: pick an option, then read the explanation.
Asked in SSC JE 01/03/2017 (Evening)
Detailed Explanation
Disputed key: the book's key is (c) 6 N/mm, found by dividing energy by extension (30/5). That gives the average force during loading in newtons, not the stiffness. Using U = ½kδ² the stiffness is 2.4 N/mm; the printed option (a) was 2 N/mm and is shown here as 2.4 N/mm.
For a gradually applied load the load–extension line is straight, so the stored energy is the triangle under it.
\[ U = \tfrac12 W\delta = \tfrac12 k\delta^2 \;\Rightarrow\; 30 = \tfrac12\,k\,(5)^2 \;\Rightarrow\; k = \frac{60}{25} = 2.4\ \text{N/mm} \qquad (W = k\delta = 12\ \text{N}) \]
Why “2.4 N/mm” Is Correct
The stiffness is 2.4 N/mm. Dividing energy by extension (30/5 = 6) gives a force (6 N, the mean load), not a stiffness.
Why the Other Options Are Wrong
- B. 4 N/mm: 4 N/mm would store ½ × 4 × 25 = 50 N·mm, not 30.
- C. 6 N/mm: 6 N/mm (book key) is energy ÷ extension — a force, not a stiffness; it would store 75 N·mm.
- D. 10 N/mm: 10 N/mm would store 125 N·mm.
Key Point
Spring energy U = ½kδ², so k = 2U/δ².
Key Concepts for Students
- Load at 5 mm: W = 2U/δ = 12 N.
- Energy grows with the square of deflection.
- Book key (c) 6 N/mm is flagged as disputed on this page.
Practice more: this is question 2 in Spring MCQs – Part 1. Try the whole set and check every answer instantly.