Quote by nrqed
Just out of curiosity: what definition of average do you use? If one uses a time average [itex] = \frac{1}{T} \int_0^T v_x dt [/itex] it does nto seem to me that the average will always be v/2.

Well, you are right, generally.
But consider this when the mass is subject to only deceleration. For a reduction in speed from v0 to v1 over an arbitrarily short time interval the change in momentum is m(v1v0) and the average speed is arbitrarily close to (v1+v0)/2. If you combine n of these intervals:
[tex]E = \sum_{j=0}^n m(v_{j+1}  v_j)(v_{j+1} + v_j)/2 = \frac{m}{2}\sum_{j=0}^n (v_j^2  v_{j+1}^2) = = \frac{m}{2}\left((v_0^2  v_{1}^2) + (v_1^2  v_2^2).... + (v_{n2}^2  v_{n1}^2) + (v_{n1}^2  v_n^2)\right)[/tex]
But the intermediate terms all cancel out and you are left with:
[tex]E = m(v_0^2v_{n}^2)/2 = m(v_n  v_0)(v_n + v0)/2 = \Delta p(v_0+v_n)/2[/tex]
If v_n = 0 then:
[tex]W = m(v_0)(v_0/2) = \Delta p(v_0)/2[/tex]
So, to be perfectly correct, one should say that the work done is the change in momentum multiplied by the average of the beginning and end speeds ([itex]W = \Delta p (v_i + v_f)/2[/itex]).
AM