It is well known that a raindrop falls under the influence of the downward gravitational force and the opposing resistive force. The latter is known to be proportional to the speed of the drop but is otherwise undetermined.Consider a drop of mass `1.00` g falling from a hwight `1.00` km .It hits the ground with a speed of `50.0 ms^(-1)` (a) What is the work done gravitational force ? (b) What is the work done by the unknown resistive force ?

Explanation

Suppose the raindrop initially at rest so `v_(0)=0`, its change in kinetic energy,
`DeltaK = K-K_(0)`
` 1/2 mv^(2)-1/2 mv_(0)^(2)`
` = 1/2 m(v^(2)-v_(0)^(2))`
`1/2 mv^(2)-1/2 mv_(0)^(2)`
` 1/2 m(v^(2)-v_(0)^(2))`
` =1/2 xx10^(-3) [(50)^(2)+0^(2)] `
` = 1.25 J `
(a) Work done by gravitational force ,
`W_(g) = mgh `
`= 10^(-3) xx10xx10^(3)`
= `10.0` J
(b) From the work energy theorem ,
`DeltaK =W_(g) =W_(g) `
` = 1.25 -10.0`
` = - 8.75 J `
The work done by resistive force is negative so displacement of rain drop will undergoes opposite the resistive force .