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

  6.8 J (nearest tenth)

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50 meters per 5 seconds