Question

Calculate the work that the friction force (air resistance) does on a ball over a vertical fall.

Heigh which the ball is dropped from: 800m

mass of ball: 0.046kg

Friction force is defined as:

$f=k\cdot v^2$

$k=4.4\cdot 10^{-4}$

Answer 1

The initial condition:

distance covered by the ball = 0m
Force on the ball = $mg$ ,
m is the mass of the ball.

Lets imagine an arbitrary position at distance s from the release point with velocity v.

acceleration,

$\large a = g - \frac{f}{m}\\ \therefore v\frac{\mathrm{d} v}{\mathrm{d} s} = g - \frac{f}{m}\\ \Rightarrow \frac{v*dv}{g - \frac{kv^2}{m}} = ds\\ \Rightarrow \int_{0}^{v}\frac{v*dv}{g - \frac{kv^2}{m}} = \int_{0}^{s}ds \ (\textup{When v =0 s = 0, v = v then s = s} \\\textup{also substituting }u = g - \frac{kv^2}{m} \textup{and integrating} )\\ \Rightarrow \frac{m}{2k}(\log_e g - \log_e(g - \frac{kv^2}{m})) = s\\ \Rightarrow \frac{g}{g - \frac{kv^2}{m}} = e^{\frac{2ks}{m}}$

We put in s = 800m and
Putting in all the values in this equation we get.

$v^2 = 1025.59\\ \therefore \frac{1}{2}mv^2 = 23.5J$ - this is the final kinetic energy.

Initial potential energy,

$mgh = 0.046*9.81*800 = 361.0J$

So work done by the friction:

$W = 23.5-361.0 = -337.5J$
(near about 340J)