A homogeneous ball weighing 2 kg lies at the bottom of a vessel with water, which rises vertically upward

A homogeneous ball weighing 2 kg lies at the bottom of a vessel with water, which rises vertically upward with an acceleration of 2 m / s ^ 2. With what force does he press on the bottom of the vessel? The density of the ball material is 4000 kg / m ^ 3. g = 10 m / s ^ 2

Answer: The ball’s gravity will be m * (g + a).
Archimedes force p * (g + a) * V, p (water density).
Total strength will
F = m (g + a) – p (g + a) V.
V = m / p1
F = (g + a) * (m – mp / p1)
F = m * (g + a) * (1 – p / p1)
F = 2 * 12 * 0.75 = 18 N
18 N total force