Help me with Physics please!!!??

Question

A spring has a force constant of 570.0 N/m.

(a) Determine the potential energy stored in the spring when the spring is stretched 4.12 cm from equilibrium.

(b) Determine the potential energy stored in the spring when the spring is stretched 2.82 cm from equilibrium.

(c) Determine the potential energy stored in the spring when the spring is unstretched.

Answer 1

a) Pes=0.5 kx^2
Pes= 0.5 x 570.0 (4.12 E-2)^2
Pes= 0.484 J

b)Pes=0.5 kx^2
Pes= 0.5 x 570.0 (2.82E-2)^2
Pes= 0.227 J

c) Still Pes=0.5 kx^2 where x=0
so Pes=0

Answer 2

E.P.E= 1/2 x Force(F) x Extension (x)
Force(F) = spring constant(k) x Extension (x)
E.P.E= 1/2 x K x x²

a) E.P.E= 1/2 x (570.0)x(0.0412)²=0.4837704 J

b) E.P.E= 1/2 x (570.0)x(0.0282)²=0.2266434 J

c) E.P.E = 1/2 x (570.0) x (0)² = 0 J