#### Answer

The amount of potential energy that was initially stored in the spring was 119 J.

#### Work Step by Step

$U_A + W_f = K_B+U_B$
$U_A = K_B+U_B - W_f$
$U_A = \frac{1}{2}mv^2+mgh - (-mg~\cos(\theta)~\mu_k~d)$
$U_A = \frac{1}{2}mv^2+mgd~\sin(\theta) +mg~cos(\theta)~\mu_k~d$
$U_A = \frac{1}{2}(1.50~kg)(7.00~m/s)^2+(1.50~kg)(9.80~m/s^2)(6.00~m)~\sin(30.0^{\circ}) +(1.50~kg)(9.80~m/s^2)~\cos(30.0^{\circ})(0.50)(6.00~m)$
$U_A = 119~J$
The amount of potential energy that was initially stored in the spring was 119 J.