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 , it attains its maximum (terminal) speed, and moves with uniform speed thereafter .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 `1 ms^(-1)` ?

Explanation

Work done by gravitational force during first half height ,
W=mgh
[ Where , `m =(rho) (4/3pir^(3))(g)(h)`
`10^(3) xx4/3 xx3.14 xx(2xx10^(-3))^(3)xx9.8 xx250`
`= 82058.66 xx10^(-6)`
` approx 0.082 J`
The change in kinetic energy of drop during first half of motion,
`DeltaK= 1/2mv^(2) -1/2 mv_(0)^(2) -1/2mv_(0)^(2) " " v=10ms^(-1) , v = 0 `
` m = rho xx4/3 pir^(3)=3.35 xx10^(-5)`
`=1/2 xx3.35 xx10^(-5) xx(10)^(2)`
`= 0.00167 J`
Work done by gravitational force during whole motion ,
= mgh
` = 3.35 xx10^(-3) xx9.8 xx500`
`=0.16415 J`
Work done by the resistive forces during whole motion = change in kinetic motion - work done by gravitatonal force
` = 0.001675 -0.1645`
`= - 0.163 J` .