Understanding derivation of kinetic energy from impulse?by richsmith Tags: derivation, energy, impulse, kinetic 

#1
May2409, 03:00 AM

P: 4

I have a question that annoys my basic understanding of kinetic energy.
I know if I have a forcetime plot then the area under the curve is equivalent to the impulse imparted on an object (in units NewtonSeconds). 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? Thanks. 



#2
May2409, 07:00 AM

Mentor
P: 40,872

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? 



#3
May2409, 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. Richard. 



#4
May2409, 10:03 PM

P: 730

Understanding derivation of kinetic energy from impulse?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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