# A Couple of Basic Questions about Mechanics

I'm writing a paper on a propulsion device and I could use some F/P ratio comparisons, like typical values for Newtons per Watt for propeller driven, jet driven, and rocket driven craft.

Also I ran in to a confusing point involving work and kinetic energy. if kinetic energy is 1/2 m v^2, then say a 1kg mass that is traveling at 1m/s, then according to the formula the mass has a kinetic energy of 0.5 joules.

Now let say the mass was accelerated from a velocity of zero to the velocity of 1m/s, by a force of 1 newton applied for one second. The definition of work state that the work applied to a mass increases the kinetic energy of that mass. So the 1 newton force applied for one second, resulted mass moving one 1 meter. The formula for work states that force times distance equals the energy added to the system, Thus 1 newton times 1 meter should equal one joule of energy added to the mass.

This is how I get two different answers for the kinetic energy of a 1kg mass that is moving at 1m/s.

Could someone point out my mistake in this please.

Best regards,

FredB

Your mistake is that a 1 N force applied for one second means that the body travels one meter. Your "jerk" (rate of change of acceleration) for 0 < t < 1, is 1. So integrating that 3 times you get:
$$s(t) = \frac{x^3}{3}$$ for 0 < t < 1

So the displacement is s(1) = $\frac{1}{3}$
Now an easier way to get work from this:
$$Work = \delta E_k$$

$$v(t) = \frac{x^2}{2}$$

this should clear things up.

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I'm still confused.

Maybe could you point me to a good article on the subject. Wikipedia is what has confused me.

Then again maybe I can just ignore the N*m approach and use the delta Ek method instead.

I could still use some ball park figures for F/P ratios for propeller planes. jets, and rockets...

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All you need to know is that $a(t) = v'(t) = s''(t)$ and that $s(t) = \int v(t) dt = \int \int a(t) dt^2$.

And for work, there are a couple of formulas you can use.

$$W = \int F \cdot dx = \int F \cdot v dt$$
$$W = \Delta E_{k}$$

In terms of a good article, I know of none but i'll look. Best description i've ever seen of work is in the Feynman Lectures - Volume 1. (Maybe because I'm a huge sucker for Feynman :P)

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Your mistake is that a 1 N force applied for one second means that the body travels one meter. Your "jerk" (rate of change of velocity) for 0 < t < 1, is 1. So integrating that 3 times you get:
The rate of change of velocity is called "acceleration". Jerk is the rate of change of acceleration.
In the example described in the OP there is a constant acceleration 1m/s^2.
The distance traveled in 1 s is 0.5 m and the work is 0.5 J, equal to the change in kinetic energy.

To the OP: just look up uniform accelerated motion.

AlephZero
Homework Helper
IF you are still confused, igmore posts #2 and #4, which are also confused and/or wrong.

To repeat what #5 said, in slightly different words:

The velocity increases linearly from 0 to 1 m/s over 1 second.
So the average velocity over the second is (0 + 1)/2 = 0.5 m/s
Distance traveled = average speed x time = 0.5 meters.
The work done = force x distance = 0.5 joules.
And that is the same as the kinetic energy = $mv^2/2$.

I see.

So I erred in presuming that 1N applied to 1kg for one second would result in a displacement of 1m. It actually results in a displacement of 0.5m, and 1N acting through a distance of 1m would increase the kinetic energy by 1 joule it would just take longer than 1 second if the origninal velocity was zero.

Thanks, that's a relief. At least I know where to look now.

the calculus text I have on hand is rather poor. Barely a paragraph about the dot product of N*m. It's single variable calculus for non-science/engineering majors.

Feyman, I'm a fan. I'll see if I can find any of his basic physic lectures that were taped.

IF you are still confused, igmore posts #2 and #4, which are also confused and/or wrong.

#4 is correct. Some of #2 is incorrect.