Work done by gravity on fluid in u-tube.

AI Thread Summary
The discussion focuses on calculating the work done by gravity on fluid in a U-tube setup with two cylinders of different fluid heights. The fluid's density is 1.30 x 10^3 kg/m^3, and the heights are 0.854 m and 1.560 m. The fluids equalize at an average height, requiring the calculation of the mass of the fluid displaced and the change in gravitational potential energy. The initial calculations omitted the gravitational constant, leading to an incorrect result. Correcting this by including the gravitational constant yields the expected answer.
Addem
Messages
20
Reaction score
0

Homework Statement



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.

Homework Equations



A version of Bernoulli's equation

p_0 + \rho g h = C

Gravitational potential energy

U=mgh

Work-energy theorem

W = \Delta U

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.
 
Physics news on Phys.org
Crap I just realized in my calculation I didn't include the gravitational constant. If I just take the answer I have an multiply by 9.8 it comes out correct.
 
Thread 'Collision of a bullet on a rod-string system: query'
In this question, I have a question. I am NOT trying to solve it, but it is just a conceptual question. Consider the point on the rod, which connects the string and the rod. My question: just before and after the collision, is ANGULAR momentum CONSERVED about this point? Lets call the point which connects the string and rod as P. Why am I asking this? : it is clear from the scenario that the point of concern, which connects the string and the rod, moves in a circular path due to the string...
Back
Top