# Velocity x' as a function of x, subject to a force function

Tags:
1. Sep 22, 2015

### danmel413

1. The problem statement, all variables and given/known data
Find the velocity x' as a function of the displacement x for a particle of mass m, which starts from rest at x =0, subject to the following force function:
F0 + cx

2. Relevant equations
Fx = m d2x/dt2

ax=Fx/m=x''=v dv/dx

3. The attempt at a solution
So I'm pretty sure I'm doing this wrong, perhaps not, but the question confused me. Right now I've rearranged the equation so it looks like

x''dx=vdv

and when I integrate both sides it becomes:

v2/2 = F0x+cx2/2

And then I multiply the right side by two and square root? It seems wrong to me. Any help would be much obliged.

2. Sep 22, 2015

### DEvens

Probably the easy way to do it is this. Notice that the fact the force depends on position means you can calculate the work done between one position and another. That means you can work out the change in kinetic energy from one location to another. Which means you can work out the speed from one location to another.

So what is it that you seem to have done here? You have F = m a. But what you really want is work equals force times distance, or in integral form, $\frac{1}{2} m v^2(x) = W(x) = \int_0^x F(s) ds$.

And, happily, that looks exactly like what you get at the end. But now you can see why it is that. (I hope.)