2 Things: Work-Energy theorem and parabolic trajectories.

1. Oct 5, 2005

misogynisticfeminist

Hi,

When reading through the work-energy thread, i just got reminded of something bugging me for quite a while. I don't think this is mentioned in the work-energy thread, firstly, how can we show that,

$$\int f \bullet ds = \Delta E$$

also,

I've heard that there's a way to show that when only one force acts on a particle or something, the particle follows a parabolic trajectory. Is there a way to show this?

thanks....

2. Oct 5, 2005

HallsofIvy

If there is a constant (non-zero) force acting on something, it follows a parabolic trajectory. That's because, taking the x-axis perpendicular to the constant force, integrating a constant twice (from $\frac{d^2y}{dt^2}= \frac{F}{m}$ to $\frac{dy}{dt}= \frac{F}{m}t+ v_0$ to [itex]y= \fra{F}{2m}t^2+ v_0t+ y_0[/tex]) we get a quadratic in t while x, with 0 acceleration, is linear in t.

(There's always "only one force" on something- the net force.)

3. Oct 5, 2005

misogynisticfeminist

ohhhh ok, that helped alot, thanks..

Anyone with the work question?

4. Oct 5, 2005

Claude Bile

You can't derive the work formula, as it is a definition. You can, however justify why the definition is a good one with a little bit of logical reasoning.

Claude.

5. Oct 6, 2005

Staff: Mentor

But you can certainly derive the Work-Energy theorem. Taking the simplest case, where the direction of the net force remains fixed:
$$\int f \bullet ds = \int m a \bullet ds = m \int \frac{dv_s}{dt} ds = m \int v_s \ dv_s = \Delta (1/2 m v_s^2)$$

6. Oct 7, 2005

misogynisticfeminist

oh ok thanks, that's what i'm looking for.

: )