Question

A small block of wood with pwood = 440 kg/m3 and Vwood = 4.5 x 10-5 m3 is submerged in water (pH2O = 1000 kg/m3) to a depth of 1.5 m. The block is then released at rest and accelerates towards the surface. How high will it travel above the surface of the water? Ignore viscosity (friction) of water. Draw a FBD for full credit.

Answer 1

Weight of the wood, W = V*rho(wood)*g = 4.5 x 10^-5 x 440 x 9.8 = 0.194 N

Buoyant force, Fb =  V*rho(water)*g = 4.5 x 10^-5 x 1000 x 9.8 = 0.441 N

So, upward force on the wooden block = Fb - W = 0.441 - 0.194 = 0.247 N

So, accleration of the block, a = Force / m = 0.247 / (4.5 x 10^-5 x 440) = 12.47 m/s^2

Suppose velocity of the block at the surface of the water = v m/s

So, v = sqrt[2*a*h] = sqrt[2*12.47*1.5] = 6.12 m/s

Now, suppose the block attains a height H above the surface of the water.

so, use the expression -

v'^2 = v^2 + 2*g*H

At the maximum height, v' = 0

so, 0 = 6.12^2 - 2*9.8*H

=> H = 6.12^2 / (2*9.8) = 1.91 m

So, the block will travel a height of H = 1.91 m above the surface of the water.

FBD of this system is very easy. Show wight W of the block in the downward direction and the buoyant force Fb in the upward direction. And the resulting acceleration in the upward direction.