Find the units of measurement of constant from formula

Question

$$m\frac{dv}{dt}=mg-kv^2$$

$v=\ms^{-1}

$m=kg$

$g=ms^{-2}$

$v^2=(ms^{-1})^2

I re-arrange the formula to isolate K

$$K=-\frac{m\frac{dv}{dt}}{v^2}+\frac{mg}{v^2}$$

Sub in the units

$$K=-\frac{\left[kg\right]\left[\frac{d}{dt}ms^{-1}\right]}{\left[(ms^{-1})^2\right]}+\frac{\left[kg\right]\left[ms^{-2}\right]}{\left[(ms^{-1})^2\right]}$$

Now I know that the derivative of velocity is acceleration but when I try to find it like a normal derivative I keep getting $-ms^{-2}$ not $ms^{-2}$. i.e

$$\frac{d}{dt}ms^{-1}$$

The m is a constant so I move it out the front. In this instance t is actually s? so I am taking the derivative of $s^{-1}$ in respect to s? So like all derivatives I take the current power and place it out the front, then take 1 from the power. i.e.

$$m\cdot{}-1\cdot{}s^{-2}$$ $$=-ms^{-2}$$

Anyway, I accept that the derivative of v is actually $ms^{-2}$ and move on.

$$K=-\frac{\left[kg\right]\left[ms^{-2}\right]}{\left[(ms^{-1})^2\right]}+\frac{\left[kg\right]\left[ms^{-2}\right]}{\left[(ms^{-1})^2\right]}$$ $$=-\frac{\left[kg\right]\left[ms^{-2}\right]}{\left[(ms^{-1})^2\right]}+\frac{\left[kg\right]\left[ms^{-2}\right]}{\left[(ms^{-1})^2\right]}$$

Which all cancels to zero which isn't correct. Could someone please explain where I am going wrong with the derivative of v thing, and the rest of the working. Thanks

Answer 1

You don't need to worry about the addition. You are trying to have all three terms of your equation have the same units. The other terms do agree in their units, as you have found, so you should have $$K=\frac{\left[kg\right]\left[ms^{-2}\right]}{\left[(ms^{-1})^2\right]}=\frac {[kg]}{[m]}$$

Answer 2

For the formula to pan out unit-wise, $mg$ and $kv^2$ will need to have the same unit. The unit of $mg$ is $kg\cdot m\cdot s^{-2}$ and the unit of $v^2$ is $m^2\cdot s^{-2}$, so $k$ needs to have the unit $$ \frac{kg}{m} \text{.} $$

BTW, for the purposes of units, you can treat differential quotients just like normal quotiest, meaning the left-hand side of your equation has the unit $$ kg \cdot \frac{m\cdot s^{-1}}{s} = kg\cdot m\cdot s^{-2} \text{.} $$ This matches the unit of the right-hand side, so all is well.