Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Work and impulse

  1. Aug 23, 2009 #1
    Work is the integral of force over a certain distance, impulse the integral of a force over a certain time span. A derivative tells us how fast something is changing at a certain point, an integral tells us how much something has changed up until that point, is this generally true (an integral is how much something has changed), or is there a better view to visualise the meaning of integration?
    Kinetic energy and momentum have very similar formulas, though kinetic energy is the integral of the resulting force with respect to distance, and momentum with respect to time, is it wrong to visualise momentum the same way as kinetic energy?
  2. jcsd
  3. Aug 23, 2009 #2
    I'm not sure how to answer this question...but for the case, in which you're trying to relate kinetic energy and momentum, it is reasonable to say that both describe an object's motion, though you should note that momentum describes an inertial property of an object in motion as a vector; kinetic energy is a scalar description of an object in motion.
  4. Aug 23, 2009 #3
    They do look similar at the begening but you have to take into account a very important fact that, i guess, you've misregarded. This is that momentum is a vector and energy is not. The kinetic energy formulae comes from the one of work (the path integral of a force over a trajectory trough the space) and the Work-Kinetic Energy theoreme wich equals one to eachother (mathematically demonstrable). Momentum, though, is a magnitude difined as p=mv where v is vectorial and so p is. 2nd Newton law claims that F=dp/dt (both p and F vectors) and there you have the integral for a certain variation of p betwen instants t1 and t2, knowing F(t).

    Concernig to the interpretation of integrals and derivatives in "real" physics, I'd just say that you should get to understant the mathematical structure of the classical physics (in this case) and then realize that as long as a measurable magnitude is treated as a real valuated function, assuming all the algebra behind them, all operators (derivatives, integrals...) and theoremes (calculus main theorem...) they "use" may have their application to the physics magnitudes.

    Hope this helps, please reply any question you have about my explanation or whatever JanClaesen.
  5. Aug 23, 2009 #4


    User Avatar
    Science Advisor

    An integral isn't how much something has changed, though. For example, the integral of a constant is the constant times the interval of integration, which is non-zero (for a non-zero constant), even though nothing changed. A good example is the integral of constant velocity, which is a simple displacement, not change in velocity or something along those lines.

    I don't know, I can't think of any good description off the top of my head that has physical significant comparable to the definition you gave for the derivative.
  6. Aug 24, 2009 #5
    Thanks, that was clarifying :smile:
    Ah indeed, the constant function, the integral value over a certain x-period is relative? So if F(2) - F(1) > F(3) - F(2) 'more' has changed in the interval 1 -> 2, 'what' has changed depends on what the x and y value represent, in the case of distance and force we call it work.

    So: the work done by a force from a point x1 to x2 equals the difference in energy of the system: the system has more possibility to do work.
    The impulse of a force from t1 to t2 equals the difference in momentum of the system.
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Similar Discussions: Work and impulse
  1. Explosion and impulse (Replies: 1)

  2. Impulse graph (Replies: 1)

  3. Impulse on balls (Replies: 2)