easy spring question, quick points?

Question

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an 8 kg stone rests on a spring. the spring is compressed 10 cm by the stone. what is the spring constnat?

the stone is pushed down an additional 30 cm and released. what is the elastic potential energy of the compressed spring just before that release?

Answer 1

F=k*x
8*10=k*(0.1)

k=800 N/m

P.E.=0.5*k*(x^2)
P.E.=64 J

Answer 2

W = mg = 8*980 dynes
k = 8*980/10 dynes/cm
F =ks
dE = Fds = ksds
E-E0 = (k/2)(s^2 - s0^2
E - 0 = (8*980/20)(40^2 -0) ergs
E = 627,200 ergs