Ah, no way a diver would be that poorly weighted but anyway
.65 g/cm^3 = 650 kg/m^3
volume of diver = 100 kg/650 kg/m^3
= .154 m^3
buoyancy force on diver = .154m^3 * 1000 kg/m^3 * 9.81 = 1509 Newtons up
diver weight down = 100*9.81 = 981 N
so
force up on spring = 1509-981 = 528 Newtons
528 = 1400 * delta L
so delta L = .377 meters
A spring with k=1400N/m is firmly anchored to the bottom of a swimming pool. A diver in full SCUBA gear with total mass of 100kg and density of 0.65g/cm3 holds on to the top of the spring. The system reaches a new static equilibrium. After the print is elongated by a length del L. How do I find del L please?
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