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

  1. A.38.8 Nm−1Nm^{-1}
  2. B.60.8 Nm−1Nm^{-1}
  3. C.190.0 Nm−1Nm^{-1}
  4. D.1900.0 Nm−1Nm^{-1}
  5. E.3800.0 Nm−1Nm^{-1}

Answer: D. 1900.0 Nm−1Nm^{-1}

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:

  1. Convert the extension to metres: e = 40 mm = 0.04 m
  2. 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.