A spring with spring constant 40 N / m is attached to the ceiling, and a 5.5- c

Question

A spring with spring constant 40N/m is attached to the ceiling, and a 5.5-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?

Explanation / Answer

Total weight of cylinder = 1.5*9.8=14.7 N

let the deflection in the spring be x at steady state and so the this much height of the cylinder will be submerged in the water

Volume inside the water= pi*r^2*x

Buoyancy force acting= V*rho*g= pi*(0.0275)^2*x*1000*9.8=23.27x

Hence Spring force= Weight - buoyancy

Kx=14.7 -23.27x

=> 40x+23.27x= 14.7

=>x=0.23m

deflection is 0.23m

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