Question:
When the potential energy is 1/2kxsqare
Answer:
Consider a spring, which has a spring constant k = 7 N/m. It has been stretched 0.70 m from its equilibrium position. What is the potential energy now stored in the spring?
In such a case, we use the formula. Potential energy = ½ k x2 = ½ * 7* (0.7)2 = 3.43 J
Thus, elastic potential energy is stored in a spring that has been stretched or compressed by a distance x away from its equilibrium position. Position x = 0 must always be the position where the spring is most relaxed. Springs have their own natural "spring constants" that define how stiff they are. The letter k is used for the spring constant, and it has the units N/m
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