Hydroynamics: Stoke's Law, Find work done by drag force

[Blandongstein]

Homework Statement

A water droplet starts falling from rest from a height $h$ and a quires the terminal speed just before reaching the ground. If $g$(acceleration due to gravity) is assumed to be constant and the radius of the drop is $r$, find the work done by air drag. The density of air is $\rho_{\text{air}}$ and the density of water is $\rho_{\text{W}}$

Homework Equations

Stoke's Law: $F_{\text{drag}}=6\pi \eta r v$

The Attempt at a Solution

Let $a$ be the acceleration of the droplet. Then,

$$\begin{aligned}mg-F_{\text{upthrust}}-F_{\text{drag}} &= ma \\ ma &= mg-mg(\frac{\rho_{\text{air}}}{\rho_{\text{W}}}) -6\pi \eta rv \\ a &= g\left(1-\frac{\rho_{\text{air}}}{\rho_{\text{W}}}\right)-\frac{6\pi \eta r }{m}v \\ v\frac{dv}{ds} &=g\left(1-\frac{\rho_{\text{air}}}{\rho_{\text{W}}}\right)-\frac{6\pi \eta r }{m}v \\ ds &=\frac{v}{g\left(1-\frac{\rho_{\text{air}}}{\rho_{\text{W}}}\right)-\frac{6\pi \eta r }{m}v } dv \end{aligned}$$

The work done by the drag force is

$$\begin{aligned} dW &= -F_{\text{drag}} ds \\ &= -(6\pi \eta r v)\left[ \frac{v}{g\left(1-\frac{\rho_{\text{air}}}{\rho_{\text{W}}}\right)-\frac{6\pi \eta r }{m}v } dv\right] \\ \end{aligned}$$

Am I doing correct? The answer given in my book is

$$\frac{4}{3}\pi r^3\rho_{\text{W}}g\left[ \frac{2r^4 g}{81\eta^2}(\rho_{\text{W}}-\rho_{\text{air}})^2-h\left( 1- \frac{\rho_{\text{air}}}{\rho_{\text{W}}}\right) \right]$$

 

[anathema]

Your attempt at the solution looks correct, but it seems like you may have made a mistake in your final equation for the work done. The answer given in the book looks more complicated, but it may be taking into account other factors such as the change in velocity and acceleration of the droplet as it falls. I would suggest double-checking your calculations and equations to see if you can arrive at the same answer as the book. If you are still unsure, it would be helpful to consult with a classmate or your teacher for clarification.