Question

A small block of wood of density 0.4xx10^(3)kg//m^(3) is submerged in water at a depth of 2.9m. Find (a) the acceletation of the block towards the surface when the block is released and (b) the time for the block to reach the surface, Ignore viscosity.

Answer 1

Given that the density of the block of wood (d(w)) = 400 kg/m³ and the block is submerged in water at depth 2.9 m.

Density of water = d(l) = 1000 Kg/m³

Upthrust on the block = d(l)Vg

For the acceleration(a) of the block -

=> d(l)Vg - mg = ma

=> a = (1000 × 10 - 400 × 10)/400

=> a = 15 m/s²

a) Acceleration of the block = 15 m/s².

Let t be time in which the block reaches the surface .

=> depth = (1/2) at²

=> (2.9 × 2)/15 = t²

=> t = 0.62 seconds

b) Time taken by the block to rach the surface = 0.62 seconds .