You hold a 1.0-kg wooden block under water with your hand (𝜌wood = 600 kg/m3). Calculate the forces acting on the block if the block is in static equilibrium. How do you calculate acceleration when the block is released?

Answer 1

let g = 9.81 m/s^2
rho water = 1000 kg/m^2 = 10^3 kg/m^3

then
for Buoyancy:
volume of block = 1 kg / 600 kg/m^3 = 1.67 * 10^-3 m^3
Archimedes says buoyant force up on block = rho water * g * 1.67*10^-3
call it B = 10^3 * 9.81 * 1.67 * 10^-3 = 16.4 Newtons up
for weight
W = m g = 1 * 9.81 = 9.81 Newtons down

that is a net force of 16.4 - 9.8 = 6.6 Newtons up

Therefore to prevent acceleration I must push down with 6.6 Newtons

If I let go a = F/m = 6.6/ 1 = 6.6 m/s^2

That is what your book says. It is not true really. The block accelerates water around it when it accelerates. The water pushes back in other words. Hydrodynamicists call the effect "added mass" for short. Therefore the block mass seems greater than its mass in a vacuum or low density gas like air.

Answer 2

Vb = 1m^3/600kg * 1kg =
1/600 = 1.67*10^-3 m^3. = Vol. of block = Vol. of water displaced.

a. 1.67*10^-3m^3 * 1*10^3kg/m^3 = 1.67 kg = Mass of the water.
F = M*g = 1.67 * 9.8 = 16.4 N. = upward force of the water. = downward
force of the hand.

b. F = M*a = 16.4.
1 * a = 16.4,
a = 16.4 m/s^2.