# Work, Impulse and F(t)

1. Mar 29, 2005

### twiztidmxcn

I'm doing problems where I have f(t) (force as a function of time) graphs and f(x) (force as function of distance) graphs.

I am just having one problem, how do I find work for a force as a function of time graph?

Im not sure how work relates at all to force as a function of time. I know that the area under the graph of F(x) curve is work, but in the case of F(t), I have no clue how I would solve it.

Oh, also, I have no F(t) = whatever, I merely have a graph and am doing graphical analysis.

Any help in how to solve for work from an F(t) graph would be much appreciated.

-TwiztidMxcn

2. Mar 29, 2005

### StatusX

F=dp/dt, where p is momentum. The kinetic energy is equal to p^2/2m. So integrating F(t) over t will give momentum, from which you can get the KE. There is a slight problem though. If you just naively assume that a momentum change corresponds directly to the work done by $\Delta W = (\Delta p)^2/2m$ you'll get into trouble because this equation is not linear in p. That is, $\Delta W_{ac} = (\Delta p_{ac})^2/2m = (p_c-p_a)^2/2m[/tex] and [itex]\Delta W_{ac} = \Delta W_{ab}+\Delta W_{bc} = (p_b-p_a)^2/2m + (p_c-p_b)^2/2m[/tex] are incompatible. So you need to know the initial momentum and use: [itex]\Delta W = E_{f} - E_{0} = p_f^2/2m - p_0^2/2m$.

3. Mar 29, 2005

### twiztidmxcn

you, my friend, i thank you for your help