$\begin{array}{1 1}650000\;ft-lbs\\400000\;ft-lbs\\250000\;ft-lbs\\450000\;ft-lbs\end{array} $

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Answer : 650000 ft-lbs

We divide into two parts , the work needed to lift the coal and the work needed to lift the cable.

The work lifting the coal is easy it weighs 800lbs and is lifted 500ft,so

$W =Fs$

$\;\;\;\;=800.500$

$\;\;\;\;=400000$ ft-lbs

Now, lets find the work done lifting the cablel; let $x$ be the height from the shaft's bottom.

A piece of cable,at initial position $x$ and of length $dx$,must be lifted $500-x$ feet.The force on pieceof cable due to gravity is $dF=2.dx$,so the work done to lift it is

$dW=2dx.(500-x)$.

$W=\int dW$

$\;\;\;\;=\int\limits_0^{500} 2(500-x) dx$

$\;\;\;\;=2(500x-\large\frac{1}{2}$$x^2)\big|_0^{500}$

$\;\;\;\;=250000$ ft-lbs

Thus the total work done is

$W_{tot}=W_{coal}+W_{cable}$

$\;\;\;\;\;\;\;=400000+250000$

$\;\;\;\;\;\;\;=650000$ ft-lbs

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