According to my physics textbook, the formula of kinetic energy is:

$$ W = \frac{1}{2}mv^2 $$

Where $m$ is is mass of the object and $v$ is the velocity of the object. The equation is calculated from this (according to that book as well): ($a$ is acceleration, $s$ is displacement, $t$ is time and $F$ is force whose value is $ma$)

$$ \begin{align} W &= Fs \\ &= mas \\ &= ma \cdot \frac{1}{2}at^2 \end{align} $$

Here comes the problem. According to that book: ($u$ is initial velocity)

$$ s = ut+\frac{1}{2}at^2 $$

But in the equation of kinetic energy, $s$ is replaced by $\frac{1}{2}at^2$, which is only possible when the initial velocity ($u$) is zero ($s = ut+\frac{1}{2}at^2 = 0t+\frac{1}{2}at^2 = \frac{1}{2}at^2$). Why the initial velocity is assumed to be zero here? What if the object has a non-zero initial velocity?


2 Answers 2


Formula $W=Fs$ is the formula for work, that is the energy transferred to the body by the force. It is the difference of kinetic energies between the final and initial state. If you're starting from $v=0$ and assume $E_{kin}(0)=0$, you have

$$ E_{kin}(v) = E_{kin}(0) + ma \cdot \frac12at^2 = 0 + \frac12 mv^2 $$

If you start from the velocity $u$ and change it to $v=u+\Delta u=u+at$, you have

\begin{align} E_{kin}(v) &= E_{kin}(u+\Delta u) = \\ &= E_{kin}(u) + ma \cdot (ut+ \frac12 at^2) \\ &= \frac12mu^2 + mu \cdot at + \frac12 m (at)^2 \\ &= \frac12mu^2 + mu\Delta u + \frac12 m (\Delta u)^2 \\ &= \frac12 m(u+\Delta u)^2 = \\ &= \frac12 mv^2 \end{align}

so the formula still holds.

  • 3
    $\begingroup$ +1. Point being, KE is the work that would have been required to bring the object up to speed if it had started from rest, regardless of whether it actually did. $\endgroup$
    – Ben51
    Commented Feb 10, 2021 at 13:45
  • $\begingroup$ How $u + v = u + at$? Is the initial velocity is zero again? $\endgroup$ Commented Feb 10, 2021 at 17:54
  • $\begingroup$ @AkibAzmain No, the initial velocity is $u$ and you accelerate by $at$. I've changed the notation abit, maybe it's a bit clearer now. $\endgroup$ Commented Feb 10, 2021 at 18:07
  • $\begingroup$ @Ben51 -- alternatively and possibly more immediately relevant in many cases: the KE you observe in your reference frame is the work that the moving object could do on something else, like if it collided. $\endgroup$ Commented Feb 10, 2021 at 18:49

The $s$ in equation $W=mas$ can be replaced from 3rd equation of motion. $$W=\frac{ma(v^2-u^2)}{2a}$$ $$W=\frac{1}{2}m(v^2-u^2) \rightarrow K.E.=\frac{1}{2}m(v^2-u^2)$$

Or else, if you still want to substitute $s$ from 2nd equation of motion:

$$W=ma(ut+\frac{1}{2}at^2)$$ $$W=ma(ut)+\frac{1}{2}m(a^2t^2)$$ From 1st equation of motion; $t=\frac{(v-u)}{a},\,\, at=v-u$ $$W=mau\cdot \frac{(v-u)}{a}+\frac{1}{2}m(v^2+u^2)-mvu$$ Rearranging and simplifying the terms we get: $$W=K.E.=\frac{1}{2}m(v^2-u^2)$$


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.