How Is Work Calculated When Pumping Muddy Water from a Barrel?

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SUMMARY

The discussion focuses on calculating the total work required to pump muddy water from a cylindrical barrel with a diameter of 1 meter and a height of 1.8 meters, filled to a depth of 1.5 meters. The density of the water varies with depth, described by the function δ(h) = 1 + kh kg/m³, where k is a positive constant. The total work done to pump the water to the top rim of the barrel is determined to be 0.366(k + 1.077)gπ joules, derived from integrating the force of gravity acting on water slices within the barrel.

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Homework Statement


A cylindrical barrel, standing upright on its circular end, contains muddy water. The top of barrel, which has diameter 1 meter, is open. Height of barrel is 1.8 meter and it is filled to a depth of 1.5 meter. Density of water at a depth of h meters below surface is given by \delta(h)=1+kh kg/m3, where k is a positive constant. Find total work done to pump muddy water to top rim of barrel. Answer is .366(k+1.077)g\pi joules


Homework Equations


force of gravity on slice=density*g*volume


The Attempt at a Solution


volume of slice\approx\pi(1/2)2\Deltah m3
1+kh(g)(\pi/4)\Deltah nt
work done on slice\approxforce*distance=1+kh(g)(\pi/4)(1.8-h)\Deltah joules
total work=\int01.5\pi/4g(1+kh)(1.8-h)dh joules=\pi/4g(1.8h-h2/2+1.8kh2/2-kh3/3)
 
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Here are some tips on LaTeX.

Use one pair of tags for a whole line, rather than piecemeal.
Inside LaTeX tags, use _n for subscripts and ^n for exponents. If the subscript or exponent is more than a single character, surround them with braces - {}
Integrals look like this (without the leading spaces in the brackets):
[ tex]\int_{a}^{b} f(h) dh [ /tex]
 

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