We know that W = 3 J and s = 99 cm = 0.99 m. From this determine k, the spring constant.W = \\[\\frac{1}{2}\\] k s2\\[\\frac{1}{2}\\] k = \\[\\frac{W}{s^2} \\]k = \\[\\frac{W}{(\\frac{1}{2} s^2)}\\]k = \\[(\\frac{3J}{[0.5 \ imes 0.99m^{2}]})\\]k = \\[\\frac{3J}{0.49m^{2}}\\]K = \\[\\frac{3 \\frac{N}{m}}{0.49m^{2}}\\]k = 6.122 N/mThank you for reading this article. We hope this article on Elastic Potential Energy was helpful to the students. If you have any doubts or suggestions then let us know in the comments section below. For more such concepts visit the website of Vedantu.