Answer
$\Delta U = 1.12~J$
Work Step by Step
We can find the elastic potential energy when the spring is compressed $4.0~cm$:
$U = \frac{1}{2}kx^2$
$U = \frac{1}{2}(3200~N/m)(0.040~m)^2$
$U = 2.56~J$
We can find the change in elastic potential energy:
$\Delta U = (2.56~J)-(1.44~J)$
$\Delta U = 1.12~J$
