How much energy can be stored in a spring with k = 470 n/m if the maximum possible stretch is 17.0 cm?

The strain energy stored in a linear spring is

SE = (1/2) * k*x²

where

k = the spring constant

x = the extension (or compression) of the spring

Given:

k = 470 N/m

x = 17.0 cm = 0.17 m

Therefore

SE = 0.5 * (470 N/m) * (0.17 m) ² = 6.7915 J

Answer: 6.8 J (nearest tenth)