Answer
$650,000$ $ft-lb$
Work Step by Step
The work needed to lift the cable is
$\lim\limits_{n \to {\infty}}{\Sigma_{i=1}^{n}}{2x_i}Δx$ = $\int_0^{500}{2x}dx$ = $[x^{2}]_0^{500}$ = $250,000$ $ft-lb$
The work needed to lift the coal is
$(800lb)(500ft)$ = $400,000$ $ft-lb$
Thus, the total work required is
$250,000+400,000$ = $650,000$ $ft-lb$
