Understanding derivation of kinetic energy from impulse?

richsmith

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?

Thanks.

Doc Al

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?

richsmith

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.

dE_logics

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.