A stretched spring has an elastic potential energy of 35 J when it is stretched 0.54 m. What is the spring constant of the spring? Round your answer to two significant figures.

The spring constant of the spring is 240.05 N/m according to the information on the question above. This problem can be solved using the spring potential energy formula which stated as PE = 1/2 * k * x^2 where PE is the spring's potential energy, k is the constant of the spring, and x is the stretched displacement. (Calculation: 240.05=35*2 / (0.54^2)

240 is the correct answer