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Home  >>  JEEMAIN and AIPMT  >>  Physics  >>  Class11  >>  Work, Power and Energy
–1 vote

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} $

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1 Answer

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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
$W =Fs$
$\;\;\;\;=400000$ 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
$W=\int dW$
$\;\;\;\;=\int\limits_0^{500} 2(500-x) dx$
$\;\;\;\;=250000$ ft-lbs
Thus the total work done is
$\;\;\;\;\;\;\;=650000$ ft-lbs


answered Aug 28, 2014 by sreemathi.v
edited Sep 11, 2014 by balaji.thirumalai

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