A vertical spring supports a cylindrical object with an outer diameter of 10 cm , a height of 20 cm , and a density of 8000 kg m ⁻³ . The object rests half dipped in water in its equilibrium state, as shown in the figure. Determine the elongation of the spring. The spring constant is 500 N m ⁻¹ .

Step-by-step solution

Volume of the cylinder, V = r^2 h = ( 0.1 2 )^2 (0.2) = 0.0005 m ^3 Mass of the cylinder, m = ho V = 8000 0.0005 = 4 kg Weight of the cylinder, W = mg = 4 10 = 40 N Since the cylinder is half dipped in water, the buoyant force is: F_B = V 2 ho_w g = 0.0005 2 1000 10 = 2.5 N At equilibrium, the upward forces balance the downward weight: kx + F_B = W 500 x + 2.5 = 40 500 x = 37.5 x = 37.5 500 m = 0.075 m x = 7.5 cm 7.5 3.14 cm = 23.55 cm 23.5 cm Answer: 23.5 cm