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As observed in the above diagram, a wooden block is held submerged in water within a container by an (external, such as a string attaching the block to the bottom of the container) force which counter-acts the upward force due to buoyancy. Clearly, if the force is terminated (for instance, by cutting the string attaching the block to the bottom of the container), the block would instantly start shooting up, while water of the same volume would fall down on account of gravity.

I would like to understand the potential and kinetic energy changes in the above system for both, the wood and the water, after the block starts moving upwards.


I am an elementary physics student and the question may sound stupid, but I am requesting help to clear my concepts. Any help would be sincerely appreciated.

How can buoyancy be explained in terms of the energy exchanges between the water and a wooden block submerged in the water withing a container, as the block moves upwards because of the force due to buoyancy?

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  • $\begingroup$ What exactly is your question? $\endgroup$
    – Bob D
    Mar 7, 2021 at 14:04
  • $\begingroup$ I want explanation of buoyancy through energy changes in the water and the wooden block individually. $\endgroup$
    – user288838
    Mar 7, 2021 at 14:23

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As we all know, the buoyancy force is given by Archimedes principle as $$ F_a = \rho g V $$ (if we direct the $y$-axis upwards), where $V$ is the volume of the object and $\rho$ is the density of the liquid.

Assuming that the surface of the reservoir is $S$, one could consider the energy of the system block + water before and after the block is submerged. Assuming that, initially, the level of the water is at $y=0$, and the block is at height $y_0=h_0$ above the water $$ E_0 = mgy_0, $$ i.e. the gravitational energy of the block. After the block is submerged to position $y_1=-h_1$, the energy becomes $$ E_1= -mgh_1 + \rho g V\frac{V}{2S}, $$ where $\rho V$ is the mass of the displaced water, $\frac{V}{S}$ is the rise of the water level, and the factor $\frac{1}{2}$ is to account that the center of mass of the displaced water is at half of its height. The first term in this equation is the gravitational potential energy of the block, whereas the second term is the potential energy of the displaced water.

Once the block fully enters water, and is acted upon by the Archimedes force, it takes work $$ W = \rho g V h_1 - mgh_1 $$ to submerge it to depth $h_1$ (this equation is approximate, as there are small corrections due to the shape of the block which account for the process where it passes through the water surface). The first part of this equation is the work done against the Archimedes force. In other words, the work done by the Archimedes force itself is $$ W_a = -\rho g V h_1. $$ Comparing the last equation with $E_1$ we see that the potential energy change is not the same as the work done - in fact, it does not depend on the depth of submersion. In other words, buoyancy force is not a conservative force (there have been contrary claims on this site, and I am open to discussion). The extra work done while submerging the object to the greater depth goes into equilibrating the water, as noted in the answer by @ClaudioSaspinski.

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When the wood block is released, the upward force $F$ is the pressure difference between its top and bottom surfaces minus the weight. That force is constant until the block reaches the surface of the water, so we can say that the potential energy is $Fh$, where $h$ is the depth of the block from the water surface.

However, the force field is not conservative, and a drag force resisting the movement is present during the path. So, the potential energy doesn't translate (completely) in kinetic energy of the block. Part of it becomes kinetic energy of the water, including waves at the surface.

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