A rain drop of radius 2 mm falls from a height of 500 m above the ground. It falls with decreasing acceleration (due to viscous resistance of the air) until at half its original height , ti attains its maximum (terminal ) speed and moves with uniform sped therefter . What is the work done by the gravitational force on the drop in the first and second half of its journey ? What is the work done by the resistive force in the entire journey if its speed on reaching the ground is `10 m s^(-1)` ?

Explanation

Here, r = 2 mm `= 2 xx 10^(-3) m`
Distance moved in each half of journey
`s = (500)/(2) = 250` m
Density of water , `rho = 10^(3) kg/m^(3)`
Mass of rain drop = Volume of drop `xx` Density
` m = (4)/(3) pi r^(3) xx rho = (4)/(3) xx (22)/(7) (2 xx 10^(-3))^(3) xx 10^(3) = 3.35 xx 10^(-5) kg`
`w = mg xx s = 3.35 xx 10^(-5) xx 9.8 xx 250 = 0.082 J`
It should be clearly understood that whether the drop moves with decreasing acceleration (or) with uniform speed, work done byt he gravitational force on the drop remains the same .
It there were no resistive forces, energy of drop on reaching the ground.
`E_(1) =mgh = 3.35 xx 10^(-5) xx 9.8 xx 500 =0.164J`
Actual energy , `E_(2) = (1)/(2) mv^(2) = (1)/(2) xx 3.35 xx 10^(-5) xx (10)^(2) = 1.675 xx 10^(-3)J`
`:. ` Workdone by the resistive forces,
`W = E_(1) - E_(2) = 0.164 -1.675 xx 10^(-3)`
W = 0.1623 Joule