1.96 kg * m2/s2
Explanation:I'm assuming that you are asking what is the elastic potential energy stored in the spring at the position stretched by 16.5 cm...Since you know the spring constant k, 144 N/m and the spring stretch from the equilibrium position x, is 16.5 cm, or 0.165, you find the spring's potential energy from the equation PE = 12kx2, which equals 1.96 Joules, or kg * m2/s2 if you want SI units.
