#### Answer

$69384~J$

#### Work Step by Step

The work required to lift the segment of the cable to the top of the can be computed as:
$W_{i} \approx (15 \Delta y) (9.8 y_i) \
J$
Therefore, the total work done on the cable when $\Delta y \to 0$ can be computed as:
$ W=\int_{0}^{12} 147 \ y \ dy \\=147 [\dfrac{y^2}{2}]_{0}^{12}\\=73.5 \times [y^2]_{0}^{12} \\=73.5 \times (12)^2 \\= 10584 \ J$
In addition, lifting the 500kg ball 12 meters requires the following amount of work:
$W=mgh=500kg*12~m*9.8m/s^2=58800~J$
Thus, the total work required is:
$58800~J+10584~J=69384~J$