Answer
a). $0.154J$
b). $0.3087J$
Work Step by Step
a). $mgsin30^{\circ}=k\Delta x$
$\Delta x=\frac{mgsin30^{\circ}}{k}=\frac{1.5\times9.8\times0.5}{175}=0.042m$
Change in elastic potential energy of system = $\frac{1}{2}k(\Delta x)^{2}=0.154J$
b). System's change in potential energy =$mg\Delta xsin30^{\circ}$
$=1.5\times9.8\times0.042\times0.5=0.3087J$
