Question

A body of mass m and density ��, which is less dense than water (�� < ��w) is dropped into a lake from a height H above the surface of the lake. To what depth H will the body sink before it begins to rise? Express your answer in terms of ��, ��w, m, and g. Ignore any loss of mechanical energy due to friction, turbulence, and water splashing.

Answer 1

The depth to which a less dense object sinks in water depends on its density in relation to the density of the water. The object stops sinking and begins to rise once the upward buoyant force equals the gravitational force. The maximum depth is then related to the object's mass, its density, the water's density, and gravitational acceleration.

Step-by-step explanation:

The question revolves around the concept of buoyancy and Archimedes' principle in physics. When an object with density ρ, which is less than the density of water ρw, is dropped into water from a height H, it will accelerate downwards due to gravity until the upward buoyant force equals the downward force of gravity. At this point, the body stops sinking and starts to rise. The buoyant force is equal to the weight of the water displaced by the submerged part of the body. We can express the maximum depth to which the body sinks before it starts to rise using the formula of buoyant force:

Buoyant force = Weight of displaced water = (Volume of body) × (Density of water) × g

Since the volume of the body × its density equals its mass, we have:

m × g = m × (ρ/ρw) × g

Therefore the maximum depth H is given by:

H = Height of fall + (m × (ρ/ρw) × g)