# 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

Staff Emeritus
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.

: )