The elastic potential energy of a spring is given by the expression [tex]PE = \frac{1}{2} kx^2[/tex], where k is the spring constant and x is the extension of spring.
So we have [tex]k = \frac{2PE}{x^2}[/tex]
PE = 1.6 Joule
x = 0.40 meter
Substituting
[tex]k = \frac{2*1.6}{0.4^2} \\ \\k =20 N/m[/tex]
So spring constant K = 20 N/m
