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Physics help?
A spring is attached to the bottom of an empty swimming pool, with the axis of the spring oriented vertically. An 7.10-kg hollow block of wood (ρ = 840 kg/m3) 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.
2 Answers
- oldschoolLv 74 years agoFavorite Answer
7.1g = k*x -----> 7.1*g/k = x
Buoyant Force = Vol*1000kg/m³ * 9.8
9800*Vol - 7.1g = k*2x
9800*Vol-kx = 2kx
Vol = 3kx/9800 = (3k*7.1g/k)/9800
Vol = 21.3*9.8/9800 = 0.0213m³
For a solid ball mass/volume = density so Volume = mass/density = 7.1/840
7.1/840 = 0.00845m³
(0.0231-0.00845)/0.0231 = 60.3%
- ?Lv 64 years ago
You don't know how much it compresses, just that it compresses an amount (call it a). So you can express spring constant k as so many Newtons per a, rather than Newtons per meter. Then when it stretches an amount of 2a, you should still be able to do the calculations. If I have time, I may come back to try to add more, to help you.
