A spring with spring constant 33N/m is attached to the ceiling, and a 4.8-cm-dia

Question

A spring with spring constant 33N/m is attached to the ceiling, and a 4.8-cm-diameter, 1.5kg 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?
y=?m

Explanation / Answer

The cylinder weight is finally balanced by bouyant force and springforce Let 'x' be the extension of spring at equilibrium, hence the height immersed in water will also be 'x' by using force balance weight of cylinder = bouyant force + spring force mg = density of water * volume immersed * g + kx where g is acceleration due to gravity = 9.8 m/s2 Density of water is 1000 kg/m3 Volume immersed = area of base * height = pi*(0.024*0.024) * x => 1.5 * 9.8 = 1000*pi*(0.024*0.024) * x * g + 33*x = 50.72 * x x = 0.2898 m

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