NECO 2009 Physics · Question 16
The work done in extending a spring by 40 mm is 1.52J. Calculate the elastic constant of the spring.
Options
- A.38.8
- B.60.8
- C.190.0
- D.1900.0
- E.3800.0
Answer: D. 1900.0
Explanation written with AI and checked against the answer key
Analyze the question: We are given the work done in stretching a spring (1.52 J) and the extension (40 mm). We need the spring constant k.
Concept: The work done in stretching a spring is stored as elastic potential energy:
W = ½ k e²
so
k = 2W / e²
Solution:
- Convert the extension to metres: e = 40 mm = 0.04 m
- Substitute the values:
k = (2 × 1.52) / (0.04)²
k = 3.04 / 0.0016
k = 1900 N/m
Tip: Always convert mm or cm to metres before using the formula. A common mistake is using 40 instead of 0.04, which gives a wrong answer like 0.38 N/m. Also remember that the ½ is part of the formula. If you forget it (k = W/e²), you get 950 N/m, which is not an option, so it is a good check.
AI can make mistakes. Report anything that looks wrong.
Past questions © the examination body (JAMB/WAEC/NECO). DeeepSolve is not affiliated with them. Rights holders: see /copyright.
