The elastic potential energy of a spring is given by
[tex]U= \frac{1}{2} k x^2 [/tex]
where k is the spring's constant and x is the displacement with respect to its unstretched position.
The initial potential energy of the spring is zero, because it is in unstretched postion, therefore x=0. The final potential energy is instead, using [tex]k=44 N/m[/tex] and [tex]x=0.14 m[/tex],
[tex]U_f = \frac{1}{2} (44 N/m)(0.14 m)^2 = 0.43 J[/tex]
And so, the increase in potential energy is
[tex]\Delta U=U_f - U_i = 0.43 J-0 J=0.43 J[/tex]
