A cable weighing 2lbs/ft is used to lift 800 lbs of coal up a mineshaft 500 ft deep.Find out how much work is done in ft-lbs.

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

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

edited Sep 11, 2014