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 P: 44 Here’s my version of the derivation of the kinetic energy formula (first post using latex, so go easy) Kinetic energy is force times change in distance. $$KE = F \Delta x$$ Force is mass times acceleration, which is a change in velocity over time. We can write this as $$F= m \frac{\Delta v}{\Delta t}$$ Now the change in distance, $$\Delta x$$ can be expressed as velocity times time $$\Delta x = \bar{v} t$$ Notice that the velocity is is average velocity, where we have a constant change in velocity, or acceleration. Average velocity can be written as the sum of the initial and final velocities divided by two. $$\bar v = \frac{v_0 + v_f}{2}$$ Now let's look at final velocity. We can write this as the initial velocity plus the change in velocity over time multiplied by time $$v_f = v_0 + \frac{\Delta v}{\Delta t} \Delta t$$ which can also be written as $$v_f = v_0 + at$$ Plugging this into the aveage velocity formula gives $$\bar v = \frac{v_0 + \left[ v_0 + \frac{\Delta v}{\Delta t} \Delta t \right]}{2}$$ which simplifies to $$\bar v = v_0 + \frac{1}{2} \frac{\Delta v}{\Delta t} \Delta t$$ Now, we plug this into the change in distance formula $$\Delta x = \left[v_0 + \frac{1}{2} \frac{\Delta v}{\Delta t} \Delta t \right] \Delta t$$ $$= v_0 \Delta t + \frac{1}{2} \frac{\Delta v}{\Delta t} \Delta t^2$$ Now, let's assume the beginning velocity is zero. Then we take the product of the above and force $$F\Delta x = \left( m\frac{\Delta v}{\Delta t} \right) \left( \frac{1}{2} \frac{\Delta v}{\Delta t} \Delta t \Delta t \right)$$ Collecting terms we arrive at $$KE = m \frac{1}{2} \Delta v \Delta v$$ Which of course is $$KE = \frac{1}{2} mv^2$$