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Understanding derivation of kinetic energy from impulse?

by richsmith
Tags: derivation, energy, impulse, kinetic
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May24-09, 03:00 AM
P: 4
I have a question that annoys my basic understanding of kinetic energy.

I know if I have a force-time plot then the area under the curve is equivalent to the impulse imparted on an object (in units Newton-Seconds). I know that this is also equivalent to the change in momentum of the object i.e. Ft = delta mV

I know that i can get these values from the plot by simply intgrating the Force function wrt to time. Now I want to determine the kinetic energy involved in this event. I know that k.e. = 1/2mV^2 so I know that is simply the integral of the impulse w.r.t. to velocity

Now that is the part conceptually I don't really grasp. What does it really mean to integrate with respect to velocity? This means I am suddenly on a velocity domain, not time, and I don't really understand this?

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Doc Al
May24-09, 07:00 AM
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P: 41,449
While impulse is ∫Fdt, work is ∫Fdx. Since force can be viewed as the rate of change of momentum, you have F = dp/dt = m dv/dt. Thus:
W = ∫Fdx = ∫m (dv/dt) * dx = m∫dv*(dx/dt) = m∫v dv = ½mv² (which is kinetic energy).

Make sense?
May24-09, 07:17 AM
P: 4
After some head scratching yes it does, finaly.

I think my mistake has always been trying to integrate wrt time. I really had to step back and ask myself what i was after, work that is, which is not a time integral.

Manipulating dv/dt to dx/dt and then subbing in V now makes it algebraicly understandable.

Thanks alot.


May24-09, 10:03 PM
P: 735
Understanding derivation of kinetic energy from impulse?

I have a question that annoys my basic understanding of kinetic energy.
Similar to questions which annoys my basic understanding of electrostatics?

Why not just calculate the final velocity of the body and then calculate the K.E.

Take as a caveat that the momentum of a body might not be directly proportional to K.E possesed by it.

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