Interactive Learning Ware 11.1 provides a review of the concepts that are import

Question

Interactive Learning Ware 11.1 provides a review of the concepts that are important in this problem. A spring is attached to the bottom of an empty swimming pool, with the axis of the spring oriented vertically. An 7.50-kg hollow block of wood (rho = 840 kg/m^3) is fixed to the top of the spring and compresses it. Then the pool is filled with water, completely covering the block. The spring is now observed to be stretched twice as much as it had been compressed. Determine the percentage of the block's total volume that is hollow. Ignore any air in the hollow space.

Explanation / Answer

Here, when the spring is in compressed condition, the force it exerts equals to the objects weight

Means -
kx = mg
k = mg/x

Now, when the spring is stretched let's look at the forces. This time for the spring the stretching we will will use 2x instead of x because the stretch is twice the compression. I shall call the buoyant force F

-2kx - mg + F = 0
=> F = mg + 2kx
Again, using our equation for k in the first part of the problem
F = mg + 2(mg/x)x = 3mg

Now by Archimedes Principle, the buoyant force equal to weight of fluid displaced. For convenience sake we shall use m for mass of block and M for mass of displaced water
F = 3mg = Mg
3mg = Mg
=> 3mg = Vg
=> V = 3m/ = 3(7.50)/(1000)
Remeber that volume of displaced water is volume of the block. Also is density of water and m is mass of block

If the block had no vacuum spots, it's volume would be
V = m/ = (7.50)/(840)
Hence the amount of empty space is the actual volume minus the theoretical volume (the volume of the solid mass of block alone without spaces)

Therefore -
Hollowy Volume = 3(7.50)/(1000) - (7.50)/(840) = 0.0225 - 0.00893 = 0.01357

So, percentage of hollow volume = 0.01357 x 100 = 13.57 %

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