Question

A stone of relative density K is released from rest on the surface of a lake.It viscouse effects are ignored, the stone sinks in water with an acceleration of:​

Answer 1

• The stone sinks in water with an acceleration of a= g(1-$\frac{1}{k}$)
• Force acting in the downward direction is Vσg
• Force acting in the upward direction is Vρg

The stone is moving downward so the net force acting on the stone in downward direction is

F = Vσg- Vρg

If m is the mass of the stone then we get,

Vσg- Vρg= Vσg ( 1- ρ/σ)

= mg ( 1- ρ/σ)

= mg (1-$\frac{1}{k}$)

Thus, Acceleration of the stone is-

a= g(1-$\frac{1}{k}$)