It is well known that a rain drop 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 1.0g falling from a height of 1.00km. It hits the ground with a speed of `50.0ms^(-1)` (a) What is the work done by the gravitational force ? (b) What is the work done by the unknown resistive force ?

Explanation

The change in kinetic energy of the drop is
`DeltaK = 1/2mv^(2)-0`
=`1/2xx 10^(-3) xx 50 xx 50)` = 1.25 J
Where we have assumed that the drop is initialy at rest.
Assuming that g is a constant with a value `10m//s^(2)`, the work done by the gravitational force is,
`W_(g) = mgh`
`=10^(-3)xx 10 xx 10^(3)`
=10.0J
b)From the work-energy theorem
`DeltaK = W_(g) + W_(r)`
Where `W_(r)` is the work done by the resistive force on the raindrop. Thus
`W_(r) = DeltaK-W_(g)`
`=1.25-10`
`=-8.75J` is negative.