# Is work is equal to $mv^2$ (without $\frac{1}{2}$)?

I was trying to come up with an equation for work that doesn't include time, because I don't know time.

Here's what I did:

$$work = Fd = mad = m{v\over t}d = m{v\over\left({d\over v}\right)}d = mv^2$$

Now, $mv^2$ is familiar, from kinetic energy equation I think, but I believe that was $\frac{1}{2} [* mv^2]$.

Where did I go wrong?

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You have (almost) rediscovered the Work-energy Theorem. Done right you get $W = \Delta \left( \mathrm{KE} \right)$. Your problems are (1) assuming you start from rest which is not true in general and (2) neglecting to use the average velocity when substituting for time (a problem that goes away if you do this with calculus). –  dmckee Aug 31 '14 at 22:43

Remember how work is defined. The key word is displacement. I think that when you wrote $Fd$, you considered $d$ to define a single point, and not displacement. Now back to your problem. The work done by a net force is $$W=Fd=mad=ma{(}\frac{v_{f}^{2}-v_{i}^{2}}{2a})=\frac{mv_{f}^{2}}{2}-\frac{mv_{i}^{2}}{2}=\Delta E_{c}$$
The single diference is that I considered $d$ to be the particles displacement (i.e. $v_{f}^{2}=v_{i}^{2}+2ad$ )
Shouldn't it be $E_k$ instead of $E_c$? –  abc Sep 1 '14 at 5:22