Question

A small sphere of wood with a density ρ = 0.40 g/cm3 is held at rest well under the surface of a pool of water. The magnitude of the initial acceleration of the sphere when it is released is? The answer is 15m/s^2 but I don't know why

A block of wood of relative density 0.750 and dimensions 20.3 cm × 30.5 cm × 40.6 cm is tossed into a choppy freshwater lake. After a reasonable time, the approximate vertical dimension above the water will be ? The answer is 5.08cm

Answer 1

The net accelerating force f = ma = B - W; where m = rho V is the wood mass, rho = .4 g/cm^3, V = 4/3 pi r^3 is the volume of the displaced water and of the sphere with radius r. B = Rho g V; where Rho = 1000 kg/m^3 is the water density and B is the buoyancy force. W = mg = rho V g is the weight of the sphere. Solve for a = ? the initial acceleration you are looking for.

a = (B - W)/m = (Rho g V - rho g V)/rho V = g(Rho - rho)/rho. g = 9.81 m/sec^2, Rho = 1 kg/m^3, and rho = .4 g/cm^3 = 400 kg/m^3.

Then a = 9.81*(600)/400 = 14.72 m/sec^2. In other words, when you let go, that little guy should pop right up. A) is the answer