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JAMB Physics 2008 Question 15

Question 15

Elasticity

A spring of force constant 500 Nm−1 is compressed such that its length shortens by 5 cm. The energy stored in the spring is

  1. A.0.625 J.
  2. B.6.250 J.
  3. C.62.500 J.
  4. D.625.000 J.
Show the answer and explanation

Answer: A (0.625 J.)

Explanation

E = ½ke2 = ½ × 500 × (0.05)2 = 0.625 J.