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Work done by force

  1. Jun 8, 2006 #1
    work done by a force

    who has derived the work done as the force x distance

    i can say that it is force x time also (of course the second one is given the name impulse) but i want to know the basics
     
  2. jcsd
  3. Jun 9, 2006 #2

    Andrew Mason

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    Force x distance is a measure of another quantity. It is not immediately obvious that this quantity is useful. But it is. Can find some reasons that force x distance is a useful quantity?

    AM
     
  4. Jun 9, 2006 #3
    Hello Chandran,

    This is a very good question you have asked. At first glance, many people will say to you "because it is defined as force times distance;" however, this is not the answer you are looking for. Essentially, you are asking for a history question. You are asking why this choice of definition, and I like it.

    I will refer you to this: http://en.wikipedia.org/wiki/Conservation_of_energy

    Read the part "Historical development"

    The answer to your question deals directly with Joules experiment, shown below:

    [​IMG]

    Looking at this picture, we can see that a known mass (which has a known force due to gravity) is displaced a vertical distance z.

    This displacement causes the paddle wheel to rotate and stir a fluid inside an insulated container (thus no heat transfer between the system [the fluid] and the surroundings [everything outside the container])

    The explanation of why the paddle wheel turns is relatively trivial. The rope is tied around the shaft of the paddle, and provides a torque as the mass falls and induces rotation.

    Now for the subtle and tricky part, why and how does the temperature of the fluid increase? Clearly, it is not because the surroundings is making it hotter (It’s insulated remember). Also, the container is not making it hotter, or we would see the container get hotter as well if they are in equilibrium.

    This means that the falling mass is responsible for the rise in temperature. If we take careful measurements of the temperature of the mass as it falls, we would find that it remains at room temperature. So we can rule out that it is not temperature itself that is being exchanged, something else must be at play here.

    One possible avenue to explore is a new concept, ‘energy.’

    At first glance, we would notice that the mass falls a distance z. So we would use this as our definition of Energy.

    By constructing careful experiments and looking at the data, we will find that our assumption of a force x distance is the correct definition to use for energy. (but not the only definition!)

    In fact, we could now test this hypothesis by rasing the mass alone to any arbitrary height z, and then letting it drop. We would find that the energy due to the displacement from the drop height to any lower point is always and exactly equal to
    1/2 *m*v^2
     
    Last edited: Jun 9, 2006
  5. Jun 9, 2006 #4

    rcgldr

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    work is defined as force x disance or as the effort it takes to create a change in energy.

    I'm not sure if there's a consensus on defining work as a change in energy and then determining that this is the equivalent of force x distance or vice versa.

    force x time is lacking information, specifically the speed at the point of application. force x speed x time = work, but since speed x time is the same as distance, you end up with force x distance again.
     
  6. Jun 9, 2006 #5
    work done by an agent

    i want to derive the formula for work done by an agent

    intuitively work done means the effort put by the agent.

    intuitively work done should be directly proportional to the force applied by the agent responsible for applying the force

    intuitively work done should also be directly proportional to the distance over which the force is applied.

    So work done=forcexdistance

    But distance=0.5*acceleration*t*t where t is the time over which the force is applied.

    since distance is proportional to time can i say

    work done=forcextime.


    what is wrong in this?
     
  7. Jun 9, 2006 #6

    Hootenanny

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    You cannot say just because something is proportional to something else that it is equal to it.

    You could also say that;

    [tex]W {\color{red}\propto} F\cdot t^{2}[/tex]

    Or

    [tex]W {\color{red}\propto} F\cdot a[/tex]

    or

    [tex]W {\color{red}\propto} F\cdot at^{2}[/tex]

    You can say any of the above. They are all proportionalities. You can also say that;

    [tex]\boxed{W {\color{blue}=} F\cdot \left(\frac{1}{2}at^2 \right)}[/tex]

    There is a big difference between proportional to and equal to.
     
    Last edited: Jun 9, 2006
  8. Jun 9, 2006 #7

    Doc Al

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    Nothing wrong with this as long as the force remains constant and there is only one force acting on the object.

    OK, assuming the object starts from rest.

    Distance is not proportional to time; it's proportional to time squared.

    Huh? What you can say (as Hoot points out), given your assumptions is:
    work = force x distance = force x (0.5*acceleration*t*t)

    OK... so what?
     
  9. Jun 9, 2006 #8
    Chandran: Work isn't defined to be force x time because you could spend all day pushing a mountain, and it doesn't move an inch. So you didn't do any "work" on it, and you spent your force on heat and sweat.

    But if you pushed a car instead, it moves. Once you've got it going, you could run round the front of it, put your hands on the hood, and find it's pushing you. Basically your pushing went into moving the car. Your "work" transferred "kinetic energy" to the car.

    You can get a feel for how much kinetic energy you gave to the car by measuring how far the car goes before it rolls to a stop. The forces stopping the car are friction and air resistance, and are fairly constant at low speeds. The distance the car rolls is maybe 50m. So if you pushed it for 5m getting faster all the while, you know that your pushing force was ten times the stopping force.

    Note that if you'd pushed the car for twice the time, it would be travelling twice as fast, but towards the end of your push you would have been almost running, and you would have covered more than twice the distance. Then when you let go of the car and let it roll to a stop, you'd find it took more than twice the distance.

    In essence, you can view "kinetic energy" as something that dictates the stopping distance of that car when subjected to a constant force. Hence kinetic energy and work are defined using force x distance.
     
    Last edited: Jun 9, 2006
  10. Jun 9, 2006 #9

    russ_watters

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    In specific cases, yes, but not in general, as Farsight pointed out with the mountain example. Or how about a book sitting on a table?

    chandran, the only way you can make time a variable is if you include the whole distance equation in the formula for work - but then time just cancels out of it anyway.
     
  11. Jun 9, 2006 #10

    Doc Al

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    Sorry if I wasn't clear enough: All my comments were directed at chandran's specific example with the assumptions stated: single, constant force acting on an object that starts moving from rest. (By "single" I mean that the force in question is the only force acting on the object.)

    I'm unclear as to what issue is troubling chandran, but--again, given the assumptions inherent in his specific example, not in general--it's perfectly OK to write work done as a function of time. (Since, trivially, the distance is a function of time.)
     
  12. Jun 9, 2006 #11

    Andrew Mason

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    Distance is not proportional to time. For an accelerating body (ie. one to which a force is applied) distance is proportional to the square of the time during which the force acts and proportional to the acceleration. If there is no acceleration, no work is done no matter how long the force is applied (in other words, if there is no acceleration there is no net force acting so no work is done).

    I think the OP is troubled with the difference between momentum and energy. Simply because there is a relationship between momentum and time (for a given force) and another relationship between momentum and energy, does not mean that these quantities are proportional.

    A 10 gram bullet moving at 500 m/sec has the same momentum as a 2.5 gram bullet moving at 2000 m/sec. But the 2.5 gram bullet will do much more damage because it has four times the energy of the 10 gram bullet.

    AM
     
    Last edited: Jun 9, 2006
  13. Jun 9, 2006 #12
    I can hazard a guess.

    Chandran, if you've got a mass m travelling at velocity v, and you exert a force upon it to bring it to a halt, the stopping distance can be calculated using ½mv² and the stopping time can be calculated using mv. One equates to the force times distance, the other equates to the force times time. We call the former kinetic energy, and the latter momentum.

    Note that if you were moving alongside the mass, in your view it would be stationary and would have no kinetic energy. The kinetic energy is in a way relative, and has no independent existence of its own.
     
  14. Jun 9, 2006 #13
    To be precise, we should be speaking of displacements and not distances when speaking of work.
     
  15. Jun 9, 2006 #14

    russ_watters

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    My spidey sense is tingling...
     
  16. Jun 9, 2006 #15

    Andrew Mason

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    The work done is the path integral of [itex]\vec{F} \cdot d\vec{s}[/itex]. A force that is applied to move an object over a surface in a complete circle does work even though its displacment is 0.

    AM
     
  17. Jun 10, 2006 #16

    rcgldr

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    Only if the force has a tangental component. The objects speed will be changed, or heat will be generated due to friction. If the only force on a object moving in a circle is a perpendicular force, then no work is done (for example an orbiting satellite).
     
  18. Jun 10, 2006 #17

    Hootenanny

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    Note the vector symbols above the force and distance ([itex]W = {\color{red}\vec{{\color{black}F}}} {\mathbf\centerdot} d{\color{red}\vec{{\color{black}s}}}[/itex]). If [itex]\vec{F}[/itex] and [itex]\vec{s}[/itex] were perpendicular then [itex]\vec{F}{\mathbf\centerdot}\vec{s} = 0[/itex] (scalar product). Hence, no work is done.
     
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