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Simple Work Formula is Parabolic with Respects to Time?

  1. Jan 26, 2010 #1
    Hello all. I'm having trouble understanding why exactly it is that the simple work formula is defined both as change in energy and force times distance, and I'm hoping that it can be made clear to me.

    Allow me to clarify. In physics, the word work has two definitions. The first, what I call the "conceptual definition", is a measure of change in energy of an object over a distance with a force acting constantly on an object. The second, what I call the "mathemetical definition", is :

    W = F x d

    with "W" being work in joules
    "F" being Force in newtons
    and "d" being distance in meters

    My problem lies in reconciling these two definitions. To me, they seem inconsistent. A force necessarily involves acceleration, which means non-linear (parabolic) increases in distance with respect to time. This, of course, means that work, being a product of what we shall assume is a constant force and a non-linear distance, is also non-linear with respect to time.

    Now, if energy is being applied through a force of constant magnitude, and we are to assume that the object is indeed moving and the force is indeed parallel to the object being acted upon, why is it that, per the work formula, energy does not flow into an object at a constant (linear) rate with respect to time? If a force is measured in newtons, and thus kilogram meters per second per second, then isn't a certain amount of energy flowing into the object per second (per any time interval, for that matter)? And isn't it the same amount of energy being transferred from second to second (or time interval to time interval, so long as the time intervals being compared are the same length)? Clearly not, though I cannot for the life of me see why. I have been laboring under what is perhaps the delusion that the magnitude of a force in newtons describes how much energy the force in question is applying per second.

    If the work formula were instead the momentum formula (p = m x v) this would make perfect sense to me. A certain speed times a certain mass yields a certain energy. If, force example, a certain amount of energy would make an object of one kilogram move at a rate of 10 meters per second, it would seem very natural to me that the same amount of energy would move an object of two kilograms at a rate of 5 meters per second. Energy proportional to both mass and velocity seems perfectly sensical, so where do we get all this stuff about force times distances and (1/2)mv^2?
     
  2. jcsd
  3. Jan 26, 2010 #2

    russ_watters

    User Avatar

    Staff: Mentor

    An applied force need not result in acceleration. It can be balanced by something like aerodynamic drag.

    Where did you get that "conceptual definition"? It isn't right, or at least, is very badly worded. There is no "change in energy of an object..." in many cases where work is performed, such as sliding a box across a floor at constant speed.
    Those units are not the units of energy, they are the units of force due to acceleration. Not all forces cause acceleration, so those units (the conversion of units via f=ma) does not always apply. If you apply a static force to a box sitting on the ground and it doesn't move, the f in Newtons is just f in Newtons.

    Anyway, you are not alone in your discomfort about kinetic energy. The easiest way to deal with it is to compare potential and kinetic energy with each other and with work. Ie,

    Work = force * distance
    Potential Energy = Weight * height

    As you can see, the above two equations are the same: Weight is a force and height is a distance.

    Now if energy is conserved, dropping a weight must result in the potential energy being converted to kinetic energy. And to do that, the weight has to accelerate at a constant rate, covering more distance (in a square function) for each next interval of time.
     
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