Why is it the case that KE = 0.5mv^2

To the best of my knowledge, energy equals work, $\mathrm{E = W}$; work equals force multiplied by distance, $\mathrm{W = Fm}$ ; force equals mass multiplied by acceleration, $\mathrm{F = MA}$; and acceleration equals distance per second squared, $\mathrm{A = m \setminus s^2}$. However, when I substitute those values in for the dimensions of $\mathrm{W}$, I can't transform the result of those substitutions into $\mathrm{W= \frac{1}{2}Mv^2}$. Perhaps I've done the algebra incorrectly.

Does the result of those substitutions transform into $\mathrm{W= \frac{1}{2}Mv^2}$?

If it doesn't transform into that equation, why doesn't it do so, and why do we use an equation that isn't compatible with our other equations?

• This is more of a physics question. – Ali Caglayan May 24 '15 at 18:26
• @Alizter I thought about that, but it's straight from my chemistry text book. Chemistry and physics overlap in a number of places, I don't see any reason why physics.se should trump chemistry.se in those cases. – Hal May 24 '15 at 18:28
• What do you get when you try it? – Ali Caglayan May 24 '15 at 18:36
• Please avoid using Latex in titles due to searching issues – bon May 24 '15 at 19:25

The kinetic energy $E_\text{k}$ equals the work $W$ done on an object: $$E_\text{k}=W$$ The work $W$ is the result of a force $F$ on a distance $s$. If $F$ is constant, the work $W$ is equal to the force $F$ multiplied by the distance $s$: $$W=Fs$$ The force $F$ on an object is equal to the mass $m$ of that object multiplied by the acceleration $a$: $$F=ma$$ For uniform acceleration $a$, the final velocity $v$ is $$v=at$$ and the distance $s$ is $$s=\frac{1}{2}vt=\frac{1}{2}at^2$$

Therefore, \begin{align} E_\text{k}&=W\\ &=Fs\\ &=mas\\ &=ma\cdot\frac{1}{2}vt=ma\cdot\frac{1}{2}at^2=\frac{1}{2}m(at)^2\\ &=\frac{1}{2}mv^2 \end{align}

@Loong's answer is perfect but I just want to show another method which can be used even if the acceleration is not uniform.

$$a=\frac{dv}{dt} = \frac{dv}{dx} \times \frac{dx}{dt}$$ Hence $$a=v\frac{dv}{dx}$$

Force is mass times acceleration.

Hence $$F=ma=mv \frac{dv}{dx}$$

For a small distance $dx$ moved, the work done by the force is $F \times dx$.

By the Work-Energy Theorem, this is also the gain in kinetic energy.

Hence $$\Delta KE = \int_{x_1}^{x_2} Fdx = \int_{x_1}^{x_2} (mv \frac{dv}{dx}) dx = \int_{v_1}^{v_2}mvdv$$

Hence $$\Delta KE = \frac{1}{2}m{v_2}^2 - \frac{1}{2}m{v_1}^2$$

If the body starts at rest $v_1=0$ and we get the kinetic energy as $\frac{1}{2}mv^2$