Physics problem about evaporation

Question

The force of gravity acts on a raindrop of mass m₀ making it fall down and slowly evaporate. Due to this, the mass of the raindrop gets smaller every second by ∆m. Calculate the force done by gravity from the beginning of the fall till the evaporation of the raindrop.

Answer 1

What a nice little problem! But I suspect that it has a typo, as you probably wanted to know the total amount of work done by gravity. I will however derive a formula for the force of gravity as a function of time, and if you average it over the lifetime of the particle you can calculate the average value of the force of gravity.

Now let me stop talking and actually solve the problem.

The mass of the raindrop falling in a vacuum as function of time t (measured in seconds) can be given by 

M(t) = M₀ - Δm t, where M₀ is the initial mass of the raindrop, assuming that the raindrop evaporates before it hits the ground.

The force of gravity at time t can be given by

F = M₀ g - Δm t g

The velocity of the particle at time t is given by v(t) = g t

The displacement of the particle dy during time interval dt is given by 

dy = v dt = g t dt 

The work done by the force of gravity during time interval dt is 

dW = (M₀ g - Δm g t) dx = (M₀ g - Δm g t) g t dt

Finally 

W = Integral from 0 to  (M₀/Δm) of (M₀ g - Δm g t) g t dt = 

M₀ g² (M₀/Δm)² / 2 -  Δm g² (M₀/Δm)³ / 3 =

M₀³ g² / (6 Δm²)

For comparison, the amount of work done by gravity on a steel ball with the same mass M₀ during the same time interval t = M₀/Δm is 3 times larger

W_ball = M₀³ g² / (2 Δm²)