Suppose that you release a small ball from rest at a depth of 0.590 m below the surface in a pool of water. If the density of the ball is 0.370 that of water and if the drag force on the ball from the water is negligible, how high above the water surface will the ball shoot as it emerges from the water? (Neglect any transfer of energy to the splashing and waves produced by the emerging ball)

Explanation:

The work of the buoyancy force will convert to gravity potential energy

Fresh Water has a density of 1000 kg/m³

The ball has a density of 370 kg/m³

assume the ball is 1 m³

weight of the ball is 370g

The buoyancy force is 1000g

Assume the buoyancy force drops suddenly to zero when the center of the ball clears the water level.

The Work done on the ball is

W = Fd = 1000g(0.590)

the change in potential energy is

PE = mgh = 370g(0.590 + y)

where y is the height above the water level.

1000g(0.590) = 370g(0.590 + y)

1000(0.590) = 370(0.590 + y)

y = (0.590)(1000 - 370) / 370

y = 1.004594... ≈ 1.00 m

no

sodium is solid at rt