A Couple of Basic Questions about Mechanics

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The discussion revolves around clarifying the relationship between work, force, and kinetic energy in physics. A participant initially misunderstands that applying a 1 Newton force for one second results in a displacement of 1 meter, when it actually results in 0.5 meters due to constant acceleration. The work done, calculated as force times distance, equals 0.5 joules, which matches the kinetic energy formula for the mass moving at 1 m/s. Participants suggest resources for further understanding, including the Feynman Lectures and a specific physics website. The conversation emphasizes the importance of correctly applying the principles of uniform accelerated motion to resolve the confusion.
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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 into 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.



FredB
 
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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 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

where your speed function is:
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&#039;(t) = s&#039;&#039;(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)

EDIT: Maybe try this page: http://physics.info/work/
 
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gordonj005 said:
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.
 
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 physics lectures that were taped.
 
AlephZero said:
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.
 
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