How do equal water depths affect gravitational potential energy?

[toforfiltum]

Homework Statement

Homework Equations

GPE=mgh

The Attempt at a Solution

I don't understand why the answer is B. My approach is assigning h a random value, say 10m. When the depths of water in both vessels are equal, the height is 5m. Therefore, since the mass of water in both vessels are still m, shouldn't the loss in GPE be half? I don't know why I'm wrong.

 

[TSny]

For an extended object, the gravitational potential energy is determined by the location of the center of gravity of the object. Consider the height of the center of gravity before and after.

 

[toforfiltum]

I still don't get one quarter. I will only get that answer if I don't consider the gain in GPE in vessel Y.

 

[TSny]

What is the initial height of the center of gravity? The final height?

 

[toforfiltum]

1/2h and 1/4h?

 

[TSny]

Yes. So, what are the initial and final values of gravitational PE?

 

[toforfiltum]

1/2mgh - 1/4mgh?

 

[TSny]

OK, that would be the loss of gravitational PE.

 

[toforfiltum]

But how about the gain in GPE in vessel Y?

 

[TSny]

But how about the gain in GPE in vessel Y?

That is already taken care of. If you consider only vessel X, then the change in PE would be

PE_f - PE_i = (m/2)g(h/4) - mg(h/2) = (1/8)mgh - (1/2)mgh = -(3/8)mgh.

Vessel Y gains PE in the amount (m/2)g(h/4) = (1/8)mgh.

Overall, there is a change of -(3/8)mgh + (1/8)mgh = -(1/4)mgh.

But you don't need to consider each vessel individually. Just consider the entire volume of water and work with the center of gravity of the entire volume.

 

[toforfiltum]

Oh, thanks. I didn't know that I should use centre of gravity to solve this question. And what do you mean by an extended object?

 

[TSny]

Extended just means that the mass is spread out over some region rather than being concentrated at a point.

 

[toforfiltum]

Ok, thanks again