1)A sphere of a given mass is put in a non viscous liquid. The sphere is released, and it moves down in the liquid. My doubt is, is the mechanical energy of the system decreasing?

The answer given in the book is no, but I dont have access to the reason. I could not think why, because, according to me, when an object is freely falling, only then is the mechanical energy conserved, but here it is experiencing an upward buoyant force, so I think this should mean that mechanical energy is not same. Is it increasing? If so, why?

2)a) The situation is the same as above, now the question is, is the gravitational potential energy of the liquid continuously increasing?

b) the liquid is now viscous, and the sphere is falling at terminal velocity. Now, is the gravitational potential energy of the liquid continuously increasing??

To for both 2a and 2b, the answer is yes. But again, I dont see how this is happening, as the level of liquid has already risen. now if the sphere moves down further, why should it have an impact on the gravitational potential energy of the liquid?

Please help!


I would guess that the phrase mechanical energy of the system means it's ability to do work.

If so, then before the sphere is released the system of the tank + liquid + sphere has some potential energy that can be harnessed to do work. As the sphere falls it's potential energy is converted to kinetic energy of the sphere and the liquid flow. This kinetic energy can still be used to do work, so the ability of the whole system to do work hasn't changed.

I note that the liquid is described as non viscous. Presumably this means that the kineti energy of liquid flow is not dissipated as heat. Note that the liquid will still provide resistance to motion because even with zero viscosity there are intertial forces. Zero viscosity just means that the liquid flows caused by the sphere's motion carry on flowing and don't come to a halt.

The question seems rather contrived.

Response to comment:

Parts 2a and b ask about the potential energy of the liquid, and presumably they mean just the liquid so you ignore the potential energy of the sphere. If you remove the sphere you leave a bubble in the liquid. This bubble starts at the top of the liquid and moves down (as the sphere falls), so when the sphere reaches the bottom you have effectively moved one sphere volume of liquid from the bottom of the tank to the top. The potential energy of the liquid has increased by the mass of liquid displaced by the sphere times the distance the centre of the sphere has moved times $g$.

The potential energy of the sphere has obviously reduced as the sphere fell, and because the sphere is denser than the liquid the net potential energy of the sphere + liquid has decreased.

The viscosity of the fluid makes no difference because the question is just asking about potential energy and that only depends on the bubble/sphere height.

  • $\begingroup$ I am confused. can you separate this into answers for 1, 2a, and 2b? $\endgroup$ – Saurabh Raje Jun 20 '13 at 14:50
  • $\begingroup$ I cant see any answer for 2a and 2b, can you help me with that? $\endgroup$ – Saurabh Raje Jun 20 '13 at 14:51
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    $\begingroup$ @Saurabh: 2a/b: If the sphere is moving down, how is the centre of mass of the liquid affected? $\endgroup$ – RedGrittyBrick Jun 20 '13 at 15:41
  • $\begingroup$ its moving down? $\endgroup$ – Saurabh Raje Jun 20 '13 at 17:02
  • $\begingroup$ @JohnRennie, 2a, and 2b is now crystal clear.. Thanks! I still didnt get the reason behind 1. You said it has kinetic energy, but what about the magnitude? Maybe, the sum of kinetic and potential at the lowest point is less than the potential energy at the highest point? As i said, this will invariably be equal if only gravity is acting, but since there is buoyancy as well, i am confused $\endgroup$ – Saurabh Raje Jun 21 '13 at 3:55

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