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Force times distance

  1. Sep 30, 2004 #1
    This one has confused me in the past. I can work with formulas and come up with the right answer... but distance relative to what? And does it mean that if I'm accelerating in a rocket then I do more and more work with every passing second if my thrust is constant? How does that make sense? Or if you look at, say, me walking from a frame of reference that is moving backwards to my direction of motion at a million miles an hour, would you come up with me doing enormously more work than looking at it from the frame of reference of the ground, because my distance covered from the million-mile-an-hour frame of reference is so much greater? If so, what do the multiple answers mean?
  2. jcsd
  3. Sep 30, 2004 #2


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    Generally relative to the point where the force was first applied.
    No, it means that the total amount of work done increases from second to second. For example, if you accelerate a mass with a force of 2 Newtons over a distance of 2 meters, you will have to expend a total of 4 Joules of work to do so.

    If you are walking at a constant speed you are doing no work. Or to be more precise, the work your muscles expend have nothing to do with moving you forward as much as they do with making up for the inefficiency of walking. The muscles have to lift you up against gravity, and alternate tension to keep you upright etc. The amount of work any muscle performs depends on how much force it exerts over how much distance. For instance if your leg has to lift you with a force of 800 N a distance of .05 meters over the course of one stride, it performs 40 joules of work.
  4. Sep 30, 2004 #3


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    You have to pick your frame first. The amount of work done will be frame dependent. But so will the energy. The kinetic energy of a mass standing still is zero. When you go to a different frame, the mass is moving, and it's energy is no longer zero. This is not a problem, energy will be conserved in any frame you select. But you do have to select one. Once you've selected your frame, you need to stay with that choice to maintain the consistency of your calculations.
  5. Oct 1, 2004 #4
    If a rocket gives a constant thrust of 1 newton in outer space then from a "starting" frame of reference the distance it goes every second will always increase. That should mean the work it does per second (the "rate of work") always increases though the force remains constant, right? If you exert a force of 2 newtons over 2 meters to do 4 joules of work, more of that work will be done over the last second (when the speed is high) than over the first second (when the object is barely moving).

    But how does that make SENSE? For conservation of energy it should require the rocket to use more energy to exert the same force force at high speeds than at low speeds, shouldn't it? But IS it? Like, a chemical rocket has a fixed amount of propellant--chemical energy--which it could use at about a constant rate for a constant thrust force. If that energy, used at a constant rate, does work (energy) at a constantly _increasing_ rate--what resolves this apparent dilemma?
  6. Oct 1, 2004 #5
    I wonder Bartholemew if you don't have a basic misunderstanding of the definition of work. Basically, work is the integral of the vector dot product of the force and a vector representation of a distance element. To make this more clear, we break the path something follows into a bunch of little same length pieces, multiply the force times the length of these little pieces and add all these products together to get the work. It doesn't matter that given a constant acceleration the rocket will travel farther each second. When you're talking about work you're talking about the total path traveled, not the path traveled per unit time; i.e the integral vs. the time derivative.

    With respect to your second paragraph, one thing you don't take into account is that as the rocket burns it's fuel, the mass of fuel onboard the rocket decreases. Thus, the rocket weighs less with each passing second, and the work is going to go down since F will go down.
  7. Oct 1, 2004 #6


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    Thet's exactly right

    You need to consider the energy in the exhaust to maintain the system energy. The detailed calculations are rather messy, but if you add up the total energy (in your fixed reference frame) of the rocket and its exhaust, you'll find that the books balance.

    For a more intuitive answer, rockets are not as energy efficient as having something to push against. For low velocities, they really suck. For high velocities, they also suck. When the final velocity of the rocket is near the exhaust velocity, they actually aren't too bad.

    The answer to your dilema at high velocities for the increasing power output of the rocket is that the rocket is burning fuel that is already moving, fuel that was accelerated by the fuel that was burned earlier. So the energy to accelerate the rocket isn't coming from nowhere, it was spent earlier, getting the rocket and it's fuel up to velocity.

    Note that the amount of fuel required to reach a given velocity grows expeonentially with the velocity target, by the well-known rocket equation

    v_rocket = v_exhaust * ln(fuelled_mass / unfuelled_mass)

    This exponential relationship is why it's very hard to reach more than a couple of times the exhaust velocity of a rocket. For very large ratios of final velocity to exhaust velocity, multistage rockets are the only way to get the mass ratios high enough.
  8. Oct 1, 2004 #7


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    It may be helpful to turn this around and consider the energy required to stop the rocket (or, for an object you deal with every day, your car). Because energy has this unusual square relationship with velocity, coming to a complete stop from 40 mph takes four times the distance as coming to a complete stop from 20 mph while you are applying the brakes for only twice the amount of time. If a deer jumps in front of you, are you concerned about the time it takes to stop or the distance? Since you are concerned about the distance, energy (or work) is what is important.

    Force times time can also be important though. So it also has a name: impulse.
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