Issue with Work-Change in KE equivalence First-year physics student, with a pretty basic question.
I've seen proofs that Work = Change in Kinetic Energy involving calculus, and they make sense to me, but I'm not sure why the following, much simpler and less general form doesn't work:
Assuming the angle between force and distance = 0, and that the force is constant:
$$W = F \cdot d = Fd = mad = m \Delta (v/t)d = m \Delta((d/t)/t)d = m \Delta (d^2/t^2) = \Delta mv^2 = 2 \Delta KE $$, which is twice what it should be. This is a very simplified case, so I'm not sure what the source of the trouble I'm having is.
 A: You are supposing that the acceleration $a$, which is the second derivative of the distance $s$ (I don't use $d$ because $\mathrm{d}d$ looks awful),
$$ a = \frac{\mathrm{d}^2s}{\mathrm{d}t^2}$$
can be written as
$$ a = \frac{\Delta v}{\Delta t} = \frac{\Delta(\frac{\Delta s}{\Delta t})}{\Delta t} = \frac{\Delta^2 s}{\Delta t^2}$$
which simply doesn't work (and $\Delta^2$ isn't properly defined, anyway).
For constant $a$, $s(t) = s_0 + v_0t + \frac{1}{2}at^2$ and $v(t) = v_0 + at$, so $a = \frac{\Delta v}{\Delta t} = \frac{a\Delta t}{\Delta t} = a$ indeed works. This is because the dependence of $v$ on $t$ is linear, so the slope of $v(t)$ is the same at every point.
On the contrary, $s$ is quadratic in $t$, and (let $s_0 = v_0 = 0$ for simplicity)
$$ v \overset{?}{=} \frac{\Delta s}{\Delta t} = \frac{\frac{a}{2}((t + \Delta t)^2 - t^2)}{\Delta t} = \frac{at\Delta t + \frac{a}{2}(\Delta t)^2}{\Delta t} = at + \frac{a}{2}\Delta t$$
is off by $\frac{a}{2}\Delta t$ from being the correct velocity. (Note that the limit $\Delta t \to 0$ would give the correct answer - this would turn the quotient by $\Delta t$ into the proper derivative w.r.t. $t$)
Thus, already the step $v = \frac{\Delta s}{\Delta t}$ is incorrect. The $\Delta$-notation (and idea) really only works if there are only linear dependences. Stick to true derivatives to avoid mistakes.
