Work done by gravity on fluid in u-tube.

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1. Dec 31, 2015

1. The problem statement, all variables and given/known data

Two cylinders of base area 4cm2 hold fluid of density 1.30(103). In one the fluid has height .854 and the other has height 1.560. They are then joined and the fluids equalize in height--find the work done by gravity on the fluid.

2. Relevant equations

A version of Bernoulli's equation

$$p_0 + \rho g h = C$$

Gravitational potential energy

$$U=mgh$$

Work-energy theorem

$$W = \Delta U$$

3. The attempt at a solution

So I know the fluids will come to rest at a height, in each cylinder, of $$\frac{1.560-.854}{2}m$$. And effectively that means taking the top chuck of fluid from the taller cylinder and placing it on top of the fluid in the shorter one. I can find the volume of this chunk of water, it's just $$\frac{1.560-.854}{2}m\times 4cm^2 \times \left(\frac{m/100}{cm}\right)^2$$. From that, together with density, I know how to find the mass.

I would think that, at this point, I just find the change in potential of the mass of fluid. The mass and gravitational constant are known. I would then just multiply by the height change, which is again $$\frac{1.560-.854}{2}m$$ and I should get the answer, no?

So I enter the following into my calculator: ((1.56-.854)/2)^2 * 4/(10^4) *1.3*10^3 and it gives me about 0.0648 when the answer in the back of the book is 0.6something.

2. Dec 31, 2015