User:Sentriclecub/Impulse work relationship

Is impulse related to work?
When an object is pushed along a surface, $$\mu_\mathrm{k}\,$$ is the ratio of the energy converted into heat vs this denominator:

$$\mu_\mathrm{k}=\frac{E_{th}}{\int F_\mathrm{n}(x) dx\,}$$

How do i verbally say the denominator? Is it the indefinite integral of the Normal Force (as a function of x) with respect to x?

I am picturing an example. A constant force is applied to block A which slides 2 meters on flat surface, then up a 60 degree incline for the final 8 meters.

I see a lot of new ways to further explore uses of $$\mu_\mathrm{k}\,$$ now, but needed to know what the denominator is in spoken words.

If this block is pulled by a string in such a way that the block has constant velocity of 1 meters per second (and the string always pulls parallel to the block's momentum), then I could integrate the normal force as a function of time, then convert it into Joules. This would involve solving for the impulse, then converting an impulse into work. I never covered subjects from Calc_II. Is it straightforward to convert an impulse to an amount of work, given this example? I have read the articles on these subjects, but they get too complicated too fast, and I get overwhelmed. I should be able to better attempt my first question if I am not afraid of making a calculus goof. Sentriclecub (talk) 13:28, 14 September 2008 (UTC)


 * For the denominator, you can just say "the integral of the normal force with respect to x" in most cases. Saying "integrating with respect to x" usually implies that every non-constant term in the integrand is somehow a function of x. You shouldn't really say "indefinite integral" in this context since, strictly speaking, the integral should have lower and upper limits (respectively, the initial and final positions of the object being moved). In the specific example you gave, it is very easy to convert the impulse into work since you know everything about the applied force -- in fact, if you use all SI units, the two quantities are numerically equal since you set the velocity to be 1 m/s. With Calc_1 knowledge, you should be able to quickly verify this using the equations $$J = \int F dt$$ and $$W = \int F dx$$ where J is the impulse and W is work. The complete answer to your question of whether there is a general relationship between impulse and work is more complicated. You definitely need to know the actual masses of the objects in question. I might explain this more in depth later, but for now I'll give you some helpful information that may lead you to work out the complete answer yourself. First, note that $$K = \frac{p^2}{2m}$$ where K is the kinetic energy and p is the momentum. Second, review the work-energy theorem, which states that the total work done on an object is equal to the object's change in kinetic energy. 97.90.132.94 (talk) 07:08, 15 September 2008 (UTC)


 * Thanks, and I hope you'll explain more, after a few days (I'll move this to my user page). I have a huge love for mathematics.  I spend additional time, when learning physics, to try under understand the equations from a extremely mathematical point of view.  I love working through proofs, and I will start working immediately on the part you gave me.  That term with momentum^2, I've never seen before, so I'll jump right on it.  I'm self taught calculus, so if I'll need more calculus knowledge to understand it, then thats a helluva motivation!  Thanks very much! Sentriclecub (talk) 14:56, 15 September 2008 (UTC)


 * The $$K = p^2/2m$$ equation is just an extrapolation from p = mv. Anyway, I should probably revise what I said earlier. As far as I can tell, there aren't any particuarly useful relationships between impulse and work (and I stress the word useful). Nevertheless, here is a helpful tip in case you haven't discovered this yet: sometimes you will need to use the equations $$J = mv - mv_0$$ and $$W_{net} = \frac{1}{2}mv^2 - \frac{1}{2}mv_0^2$$ and notice that $$v^2 - v_0^2 = (v+v_0)(v-v_0)$$. Sorry if I might have wasted your time with my earlier statement. You would still profit from learning more about the work-energy theorem, though. 97.90.132.94 (talk) 19:16, 16 September 2008 (UTC)


 * Interesting, you restated my most fascinating insight ever. Namely that $$v^2 - v_0^2 = (v+v_0)(v-v_0) = 2ad $$ and now divide both sides by two, assuming constant acceleration, the term (2) is eliminated once you see that $$ \frac{1}{2}(v+v_0) = v_{avg}$$ and realizing that $$ \Delta v = v-v_0 \,$$ and voila look what this amateur mathematician derived... (assuming constant acceleration) $$ \mathbf{a} \mathbf{d} =  \Delta v *  v_{avg} $$ and now you can truely see the relationships because now all the terms are intuitive, and if you are the type to try to make your brain see intuitive relationships that exist in mathematics, then this equation is beautiful.  Now multiply both sides by $$ \frac{m}{v_{avg}}$$... work divided by average velocity = change in momentum = impulse. Sentriclecub (talk) 11:44, 18 September 2008 (UTC)

Leave comments here, any feedback appreciated!
For instance, why didn't I see that the 1 m/s velocity is the only velocity which makes |impulse| = |work|, duh!

I´m not sure K = p2/2m has much meaning, more then just being a basic algebra trick. I think it´s more worth while noticing that dK/dv = p, that is, the derivate of kinetic energy with respect to velocity equals |p|