You stretch a spring by a distance of 0.3 m. the spring has a spring constant of 440 n/m. when you release the spring, it snaps back. what is the kinetic energy of the spring as it reaches its natural length?
a.5.94 j
b.8.22 j
c.21.5 j
d.19.8 j

The kinetic Energy has to be the same as the Energy with which the spring was stratched:
1/2×m×v^2=1/2×D×s^2
Ekin=1/2×440×0.3^2= 1/2×440×0.09=19.8. ( Of course, the kinetic Energy is normally Not calculated Like this , but in this case both Formulas have the Same solution so you can Wright it Like This)
By the way, if the spring Had Not reached its Natural length yet, you wouldnt be right if you calculate it Like above, because a Part of the Energy is still in the spring : Espring=Ekin+Espring2

Did you ever figue this