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JAMB Physics 2007 Question 30

Question 30

Elasticity

A string of length 4 m is extended by 0.02 m when a load of 0.4 kg is suspended at its end. What will be the length of the string when the applied force is 15 N? [g=10ms−2]

  1. A.5.05 m
  2. B.6.08 m
  3. C.4.05 m
  4. D.4.08 m
Show the answer and explanation

Answer: D (4.08 m)

Explanation

k = F/e = (0.4 × 10) ÷ 0.02 = 200 N m−1. A 15 N force stretches it 15 ÷ 200 = 0.075 m, so the length is 4 + 0.075 ≈ 4.08 m.