Let's say a vehicle that weighs 20t is hauling along at 50m/s and we want to brake it down to a full stop.
The kinetic energy we need to dissipate into heating up the brakes is
0.5 * 20.000kg * 50m/s * 50m/s=25MJ
Now let's say we first brake down to 25m/s and then to 0.
We dissipate
0.5 * 20.000kg * 25m/s * 25m/s=6.25MJ
in the first braking action from 50m/s to 25m/s and then another
0.5 * 20.000kg * 25m/s * 25m/s=6.25MJ
in the second braking action from 25m/s to 0m/s, thus making the brakes consume a total of 12.5MJ in heat they need to dissipate.
That's obviously completely wrong, because where did the extra 12.5MJ go? I'm missing something blatantly obvious here that makes the addition totally bogus, so how to fix that?
Best Answer
The formula for kinetic energy is $\frac12mv^2$.
If your initial velocity is $v_i$ and your final velocity is $v_f$, then your initial kinetic energy is $KE_i = \frac12 m v_i^2$ and your final kinetic energy is $KE_f = \frac12 mv_f^2$. The difference is $\Delta KE = KE_f - KE_i = \frac12 m(v_f^2 - v_i^2)$
It appears you're thinking that you can define $\Delta v = v_i - v_f$ and then set $\Delta KE = \frac12 m \Delta v^2$. That is not right. Mathematically, it is not the same as the formula we found before. Instead, it is the same as $\Delta KE = \frac12 m(v_f - v_i)^2 = \frac12 m (v_f^2 - 2v_iv_f + v_i^2)$.
So the specific problem you had was that you found the change in speed from $50 m/s$ to $25 m/s$ first and then tried to calculate the kinetic energy change from the result. Instead, you have to calculate the kinetic energies first, then subtract to find the change in kinetic energy.