If a 60-N force is compressing the spring a distance of 5 mm = 0.005 m, then by Hooke's law, the spring has constant k such that
60 N = k (0.005 m) ⇒ k = 12,000 N/m
Then the spring is storing
1/2 (12,000 N/m) (0.005 m)² = 0.15 J
of potential energy.
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