# Deriving x(t) from work-energy theorem

1. Feb 17, 2013

### autodidude

In K&K's text on mechanics, after they present the derivation of the work energy theorem:

$$\frac{1}{2}mv^2-\frac{1}{2}mv_0^2=\int_x_0^x F(x) dx$$

It is mentioned that since $$v=\frac{dx}{dt}$$, we could solve for $$\frac{dx}{dt}$$ and integrate again to find x(t)

I tried that with $$v_0=0$$ just to make things a little easier and ended up with something like:

$$x=\int \sqrt{2m(F(x)-F(x_0))}dt$$

which looks horrible and wrong

Is this heading down the wrong path? How would you do it when $$v_0$$ is not zero?

2. Feb 17, 2013

### voko

Your derivation is not right; it must be an integral of force rather than force itself under the radical.

As an example, consider the case when F = -kx (mass attached to a spring).

3. Feb 17, 2013

### haruspex

As voko says, plus, you have m where it should be 1/m.