Question

how much energy is stored in a 7.0 cm spring that has a spring constant of 500 N/m when it is stretched 3.0 cm​

Answer 1

Answer:

12250 J

Explanation:

elastic potential energy = 0.5 × spring constant × (extension)2

spring constant = 500

extension =  7cm

7^2=49

0.5 × 500 × 49 = 12250 J or 12.25kJ