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JAMB Physics 2003 Question 2

Question 2

Elasticity

On top of a spiral spring of force constant 500Nm−1 is placed a mass of 5 × 10−3 kg. If the spring is compressed downwards by a length of 0.02m and then released, calculate the height to which the mass is projected. [g = 10ms−2]

  1. A.8m
  2. B.4m
  3. C.2m
  4. D.1m
Show the answer and explanation

Answer: C (2m)

Explanation

Energy stored = ½ke2 = ½ × 500 × 0.022 = 0.1 J. This becomes mgh, so h = 0.1 ÷ (5 × 10−3 × 10) = 2 m.