Two identical cylindrical vessels with their bases at the same level each contain a liquid of density 1.30 × 10^3 kg/m^3. The area of each base is 4.00 cm^2, but in one vessel the liquid height is 0.854 m and in the other it is 1.560 m. Find the work done by the gravitational force in equalizing the levels when the two vessels are connected.

Question

amilapsn · Answer

think about the initial and final positions of the center of gravity of the liquid.

ganeshie8 · Answer

I think the difference in y coordinate of center of gravity of the displaced liquid would be :

(1.56-0.854)/2 = 0.353

ganeshie8 · Answer

perhaps it is 0.353/2

?

amilapsn · Answer

think a bit more.... drawing a  sketch would help

amilapsn · Answer

yep you're right. its .353/2

ganeshie8 · Answer

should i simply find the mass and calculate the reduced potential energy of that displaced liquid  ?

amilapsn · Answer

yep!

ganeshie8 · Answer

4/10000*1.3*1000*9.8*0.353 = 1.799 J

let me check the answer...

ganeshie8 · Answer

answer is 0.635 J

let me double check my calculations..

amilapsn · Answer

work done=displaced liquid's mass$\times$g$\times$displacement of the center of gravity

ganeshie8 · Answer

mass =  density * volume
         = 1.3 * 10^3 * (4/10000*0.353)
         = 0.184 kg

ganeshie8 · Answer

does that look good ?

amilapsn · Answer

yeah..

ganeshie8 · Answer

W = mgh = 0.184*9.8*0.353/2 = 0.318 J

still no luck...

amilapsn · Answer

wait a minute.

ganeshie8 · Answer

okie

imqwerty · Answer

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