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 height $1.00$ km. It hits the ground with a speed of $50.0 m s^{- 1}$ .What is the work done by the unknown resistive force?

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Solution

The change in kinetic energy of the drop is

$Δ K = \frac{1}{2} m v^{2}$

$= \frac{1}{2} \times \times 10^{- 3} \times 50^{2} = 1.25 J$

Assuming that g is a constant with a value $10 m / s^{2}$ .

The work done by the gravitational force is

$W_{g} = m g h$

$= 10^{- 3} \times 10 \times 10^{3} = 10.0 J$

From the work-energy theorem

$W r = Δ K - W_{g}$

$= 1.25 - 10 = - 8.75 J$ is negative.

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The change in kinetic energy of the drop isΔK=12mv2=12××10−3×502=1.25JAssuming that g is a constant with a value 10m/s2.The work done by the gravitational force isWg=mgh=10−3×10×103=10.0JFrom the work-energy theoremWr=ΔK−Wg=1.25−10=−8.75J is negative.