How can the work required to remove a solid ball from water be calculated?

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To calculate the work required to remove a solid ball from water, Archimedes' Principle is essential, as it states that the buoyant force equals the weight of the displaced water. The ball, having the same density as water, is neutrally buoyant, meaning it requires minimal force to move in any direction. However, when the ball is submerged, work is done on the water due to the increase in water level. The conservation of energy indicates that the same amount of work is needed to remove the ball from the tank. The dimensions of the tank are not specified, but its width allows for negligible changes in water level during the process.
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Homework Statement



A solid ball with radius r and mass m is located inside an opened wide tank filled with water. The ball has the same density with water.
http://3.bp.blogspot.com/_jn57XA2jL...41pJXdRc/s1600-h/Get+Ball+Out+from+Water.jpg"
Calculate work required to get the ball out from the water!
Ignore viscosity and surface tension.

Homework Equations



Fbuoy = ρgV
Fgrav = -mg

The Attempt at a Solution



I have found its answer, W = mgr, but http://collectionofphysicsproblems.blogspot.com/" , it says there's a tricky solution.
I hope anyone could help on finding the tricky solution.
Thanks.
http://collectionofphysicsproblems.blogspot.com/"
 
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Use Archimedes' Principle to figure this one. Think about the amount of water displaced when the ball is completely submerged.
 
Hmm...
Wouldn't it really need integral?
If I use archimedes buoyancy force, it would...
W=integral(ρgV dh)
 
An integration is not required. Archimedes Principle states the bouyant force is equal to the amount of fluid displaced. The ball has the same density of water so it neutrally bouyant (bouyant force equals gravitational force). So, an infinitesimal amount of force will move the ball (ignoring viscosity) in any direction. But work was done on the water when the ball was completely submerged (water level increased). Conservation of energy states the same amount of work is required to remove the ball from the tank.
 
It is given that the tank is wide, so the rise in the level of liquid is ~~ negligible.
and Neither we have dimensions for tank.
 
Kindly see the attached pdf. My attempt to solve it, is in it. I'm wondering if my solution is right. My idea is this: At any point of time, the ball may be assumed to be at an incline which is at an angle of θ(kindly see both the pics in the pdf file). The value of θ will continuously change and so will the value of friction. I'm not able to figure out, why my solution is wrong, if it is wrong .
TL;DR Summary: I came across this question from a Sri Lankan A-level textbook. Question - An ice cube with a length of 10 cm is immersed in water at 0 °C. An observer observes the ice cube from the water, and it seems to be 7.75 cm long. If the refractive index of water is 4/3, find the height of the ice cube immersed in the water. I could not understand how the apparent height of the ice cube in the water depends on the height of the ice cube immersed in the water. Does anyone have an...
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