# The concept of work and types of energy

## Main Question or Discussion Point

The concept of "work" and types of energy

I have read the other posts regarding the concept of work and still don't understand why distance has anything to do with it. Anyway, I thought that maybe an explanation of the following might help:

Suppose you have a 1kg space-ship in an empty universe, with zero velocity in some frame of reference. Then it consumes its rocket, which is able to push with 1 Newton force for 1 second**. The ship is now going 1meter / second and has a total energy of 1/2*mv^2 = 0.5 Joules kinetic energy.

Now suppose the same scenario, but the ship initially has a 1 meter / second velocity. The final velocity is 2 meters / second = 1/2*mv^2 = 2 Joules.

So in the first case, the transition of chemical energy to kinetic energy results in a gain of 0.5 Joules for the ship, whereas in the second case, a gain of 1.5 Joules occurs.

**Is it safe to say that a rocket can provide the same impulse, regardless of starting velocity?

Related Other Physics Topics News on Phys.org

You are correct. The same rocket will provide the same impulse.

Here's a simple derivation from basic kinematics that should help convince you that W = Fd. It's not a perfect derivation, because I wont factor in any possible potential energy, but here it goes:

$d = \frac{v_2^2-v_1^2}{2a}$

therefore,

$\Delta v^2 = 2ad$
$\frac{1}{2}\Delta v^2 = ad$

multiplying both sides by mass:
$\frac{1}{2}m\Delta v^2 = mad$

F = ma, and (1/2)mΔv2 = ΔE, so:

$\Delta E = Fd$

and ΔE is defined as the Work.

If you look at equations for, say, graviational potential, you'll very easily see that the change in potential energy is the work:

ΔU = mgh is the same as saying ΔE = W = mad = Fd where g→a, and h→x. This being the case where only the potential changes, and not the kinetic. This should convince you as to why distance is necessary in work calculations.

Last edited:
DrDu