Problem involving hookes law and friction

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The discussion centers on solving a physics problem involving Hooke's Law and friction, emphasizing energy conservation principles. Key equations include spring potential energy (Es = kx²/2), kinetic energy (Ek = mv²/2), and gravitational potential energy (Eg = mgh). The energy transfer from spring energy to kinetic energy is crucial for determining velocity at different points. The approach involves equating energies to find final velocities and using friction to calculate distance traveled. Overall, the focus is on utilizing energy equations for a clearer understanding of the problem.
guitarist2400
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This is a question my professor asked in class and I did not understand his explanation at all, I found your forums hoping someone would break it all down for me

[PLAIN]http://i28.lulzimg.com/0cb917f809.jpg
 
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I'm assuming you already know the answer, and you just need the methods explained more?This problems seems to be an energy problem.

The equations regarding energy are
Es = kx2/2
Ek = mv2/2
Eg = mgh
Also, the concept that E = fd is helpful.To find velocity at the top, the only energy transfer occurring is from spring energy to kinetic energy.
If you set the equations for those two equal you can solve for v.

Velocity at the bottom is almost the same process.
The transfer of energy is from the velocity at the top and from gravity to velocity at the bottom (you can also set spring energy and gravity energy equal to the kinetic energy at the bottom).
Setting Eki (or Es) + Eg = Ekf will allow you to solve for vf.

Distance the object travels with friction can be used with kinematic equations, if you solve for acceleration.
I find its easier to use energy, since you already will have solved for energy at the bottom (Ekf).
Since the friction will dissipate all of that kinetic energy before the object stops, you can use the E = fd with E being Ekf and F being uFn.
 
The book claims the answer is that all the magnitudes are the same because "the gravitational force on the penguin is the same". I'm having trouble understanding this. I thought the buoyant force was equal to the weight of the fluid displaced. Weight depends on mass which depends on density. Therefore, due to the differing densities the buoyant force will be different in each case? Is this incorrect?

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