A rain drop of radius 2 mm falls from a height of 500 m above the ground. It falls with decreasing acceleration until at half its original height, it attains its maximum 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 `10ms^(-1)`?

Explanation

`r=2xx10^(-3)m,h=500m`
Work done `W_(1)=(1)/(2)m(v_(B)^(2)-v_(A)^(2))`
where `m=rhoV=(4)/(3)pir^(3)rho`
`therefore W_(1)=(1)/(2)xx(4)/(3)xx3.142xx(2xx10^(-3))^(3)xx10^(3)xx10^(2)`
`=1.676xx10^(-3)J`
(work done) `=DeltaKE=W_(2)=(1)/(2)m(v_(c)^(2)-v_(B)^(2))`
`because" " v_(C)=v_(B),W_(2)=0`
Work done = `W_(1)+W_(2)=1.676xx10^(-3)J`