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# We have a cable that weighs 2lbs/ft attached to a bucket filled with coal that weighs 800lbs.The bucket is initially at the bottom of a 500 ft mine shaft.Determine the amount of work required to lift the bucket to the midpoint of the shaft.

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

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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
$\;\;\;\;\;\;\;=2(500-x)+800$
$\;\;\;\;\;\;\;=1800-2x$
$W=\int\limits_0^{250} F(x) dx$
$\;\;\;\;=\int\limits_0^{250} 1800-2xdx$
$\;\;\;\;=(1800x-x^2)\big|_0^{250}$
$\;\;\;\;=387500\;ft-lb$
answered Sep 1, 2014
edited Sep 2, 2014

–1 vote