$\begin{array}{1 1}387500ft-lb\\287500ft-lb\\187500ft-lb\\587500ft-lb\end{array} $

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Answer : 387500 ft-lb

Assume $x$ to be the amount of cable that has been pulled up.

So at the bottom of the shaft $x=0$, at the midpoint of the shaft $x=250$ and at the top of the shaft $x=500$

Also at any point in the shaft there is $500-x$ feet of cable still in the shaft.

The force then for any $x$ is then nothing more than the weight of the cable and bucket at that point.This is,

$F(x)=$ weight of cable +weight of bucket/coal

$\;\;\;\;\;\;\;=2(500-x)+800$

$\;\;\;\;\;\;\;=1800-2x$

$W=\int\limits_0^{250} F(x) dx$

$\;\;\;\;=\int\limits_0^{250} 1800-2xdx$

$\;\;\;\;=(1800x-x^2)\big|_0^{250}$

$\;\;\;\;=387500\;ft-lb$

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