A spring with spring constant 31 N/m is attached to the ceiling, and a 5.3-cm-di

Question

A spring with spring constant 31 N/m is attached to the ceiling, and a 5.3-cm-diameter, 1.2 kg metal cylinder is attached to its lower end. The cylinder is held so that the spring is neither stretched nor compressed, then a tank of water is placed underneath with the surface of the water just touching the bottom of the cylinder. When released, the cylinder will oscillate a few times but, damped by the water, quickly reach an equilibrium position.

when in equilibrium,what length of the cylinder is submerged?

Explanation / Answer

If we assume the cylinder is hanging such that the circular end is parallel with the water and that the tank water level is not materially affected by the displacement of the cylinder as it sinks, and that the water is fresh and has a density of 1000 kg/m³.

Let x be the spring stretch and equivalent sinking distance in meters

mg = kx + gAx
mg = x(k + gA)
x = mg / (k + gA)

x = 1.2(9.81) / (31 + 1000(9.81)((0.053)²/4)
x = 0.2236 m
or 22.36 cm of stretch or sinking

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