Answer
$\Delta U = -0.80~J$
Work Step by Step
We can find the elastic potential energy when the spring is stretched $2.0~cm$:
$U = \frac{1}{2}kx^2$
$U = \frac{1}{2}(3200~N/m)(0.020~m)^2$
$U = 0.64~J$
We can find the change in elastic potential energy:
$\Delta U = (0.64~J)-(1.44~J)$
$\Delta U = -0.80~J$
