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B The concept of a force lifting a weight and the work done

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  1. Oct 2, 2018 #1
    if I apply a force constant F=100N en ##y## direction to a body of mass 2kg, it is elevated 3m and I know that the force of gravity is in ##y## direction too. What is the work that I did?
    I think: ##W_{me}=(F_{me}-F_{g})\times 3## it is correct?
    and the work made by gravity is "0".
     
    Last edited by a moderator: Oct 3, 2018
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  3. Oct 2, 2018 #2

    jbriggs444

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    No, that is not correct. The work done by a given force over a given distance does not depend at all on what other forces happen to be acting.
     
  4. Oct 3, 2018 #3

    sophiecentaur

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    I am constantly coming across confusion about who does the work to whom. It's not fundamental and the question needn't actually arise as long as the 'rules' and conventions are followed (i.e. using the right forces and distances). It's basically a matter of where the Energy is going. In questions of the kind that the OP refers to, there can be Energy dissipated in one part of the setup and some 'useful' work done elsewhere. Lifting a mass with a machine with significant of friction is just one example. I personally wouldn't say that the Friction had done any work.
     
  5. Oct 3, 2018 #4

    jbriggs444

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    I personally would say that a force pair involved in kinetic friction had done negative work. Of course, it's just a question of bookkeeping. At the end of the day, as long as the books balance, the column names you use when you write down numbers are not very important. It's picking a particular convention and adhering to it consistently that matters. As you point out.
     
  6. Oct 3, 2018 #5

    CWatters

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    As jbriggs444 said, that's not correct.

    In the set up you describe you are applying a defined force 100N over a defined distance 3M. That's all you need to calculate the work done.

    It doesn't matter that some of the force is used to overcome gravity and some is used to accelerate the object or some used to overcome friction etc. By stating you applied 100N you are stating the total force you applied and hence the work done.

    The answer would be different had you asked how much work has been done against gravity.
     
  7. Oct 3, 2018 #6

    russ_watters

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    Are you saying you would consider energy dissipated different from work? There are some contexts where that might work out ok, but I'd be concerned about accidentally implying a violation of conservation of energy or creating some confusing limitation in a description of a scenario. E.G., if you shove a block across the floor and then release it, how do you calculate how far it slides without saying the floor is doing (negative) work on the block? Kinetic energy dissipated by friction is equal to the work it does on the block.
     
  8. Oct 3, 2018 #7

    sophiecentaur

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    You could say that the floor is doing work on the block too. Which of the two would be dissipating most energy (heat)? It would have to depend on the nature of the actual surfaces and that would be one reason why I personally find the Useful Work would be the only work to be identifiable. All the rest, in this case, would be just wasted energy.
    I am not contradicting conservation of Energy just suggesting a clearer use of the term Work. For instance, if you say that the Earth does work on the accelerating car, that could make sense becaure the car actually acquires KE. The mass ratio is so huge that the Earth gains virtually no momentum and virtually zero KE. How, then is it reasonable to say the car is doing work ON the Earth if there is none of the Fuel Energy transferred to tthe Earth? Mathematically, it's true but the description of the event is misleading. The car is really doining work on Itself! because the Energy starts and ends wih the car. That's my problem.
     
  9. Oct 3, 2018 #8

    jbriggs444

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    The amount of work done by the floor on the box is an easily calculable quantity. The amount of work done by the box on the floor is an easily calculable quantity. Both results will depend on your choice of reference frame. But there is a useful invariant. If you add up both figures, the result will be negative. That quantity does not depend on your choice of reference frame. It is the negative work done by the kinetic friction force pair (which is what I referred to several posts above).

    I do not see anyone claiming that one can tell where the lost mechanical energy will manifest as heat based on an examination of work done. Only that one can tell how much will manifest.
     
    Last edited: Oct 3, 2018
  10. Oct 3, 2018 #9

    russ_watters

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    Did you mean to reverse what I said? You just repeated it as stated. If you meant the block is doing work on the floor, it is, but in these types of problems we generally focus on one object at a time. Mixing and matching forces on different objects is what creates the confusion you referenced at the start of your first post. Keeping it straight means focusing on the block:

    1. When you push it, you apply a force over a distance, providing positive work.
    2. When friction slows it down, it applies a force over a distance, providing negative work.
    That the friction force is dissipative can be used as a way to define [not]"useful" work, sure. I like that much better than avoiding even saying it is work at all (which is what you said in your first post that you prefer).
    Momentum yes, KE, no. But regardless/again, focusing on one object at a time avoids accidentally mixing and matching.

    Where and exactly how the energy is added or dissipated doesn't have any effect on the work/energy relationship; it's just a means for transmitting the energy (unless of course its the mechanism itself that is being asked about). That's another common misunderstanding we get -- just last week we had a thread on the mechanics of walking, where it was an issue (a frequent example). To simplify, imagine a spring pushing a block away from a wall. The mass of the spring is a tiny fraction of the mass of the block. Constrain the analysis to the expansion of the spring to neutral position. Does it actually matter to the work/energy calculation whether the spring is attached to the block, the wall, both or neither? Problems arise when people do the analysis assuming the spring (your leg) is irrelevant, and then at the end decide they want it to be relevant and can't understand why their own analysis is both correct and incorrect at the same time!

    [Mechanical] Energy is not an object that gets handed from one entity to the other. It's not a cake that gets baked in one place, being created there and staying there. It's the effect of an interaction between the two objects. Sure, you can say that at its most basic, burning fuel provides the car with kinetic energy. But the how of it is due to a Newton's 3rd Law force pair interaction between the Earth and ground. Whether the fuel is burned in the car or on the ground isn't critical to the analysis: it usually just gets in the way.
     
  11. Oct 3, 2018 #10

    sophiecentaur

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    Ahh yes. I had an internal discussion and got the wrong output. I agree with what you have been saying except that saying Work is done as defined by Physics but when no energy is gained (the Earth gains no KE from the exercise), to say work has been done on it seems wrong. It's along the lines of the problem of putting electrical energy into a dead short or open circuit - the equivalent of a gross impedance mismatch.
    But the definition of Work has to stand, of course.
     
  12. Oct 3, 2018 #11

    sophiecentaur

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    The ratio of Energy transferred is proportional to the velocity ratio squared and that's more than the ratio of Momentums.
    I am onto a bit of a loser here, I know, because I can't argue with the definitions. It's just the useage of the terms that goes against simple understanding.
     
  13. Oct 5, 2018 #12

    Mister T

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    Me, too. Note that if one insists that the friction force does work, it is not possible to explain where the energy is going.
     
  14. Oct 5, 2018 #13

    Mister T

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    No.

    ##W_{me}=F_{me}\times 3##
     
  15. Oct 6, 2018 #14

    sophiecentaur

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    This was sorted out in EE where batteries have an emf in them but resistors don’t. Perhaps we should invent the ‘mf’ for ME.
     
  16. Oct 9, 2018 #15
    Friction can do work and the energy goes up in heat. a car traveling at 50mps at 1000kg has a KE value of 1.25MJ. when it stops and requires a force created by friction to slow it in x distance in y time, that x*y is the work done by the brake friction which goes up in heat and noise.
     
  17. Oct 9, 2018 #16

    Mister T

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    Where does the energy come from to create this increase in temperature?

    Imagine a block moving across a level table top at a steady speed. There's a force propelling the block that's equal and opposite to the force of friction on the block. If you think the work done by the propelling force is equal and opposite to the work done by friction, then the net work done is zero, and the change in kinetic energy is zero. This is perfectly valid from a dynamics perspective. But from an energy perspective we are left trying to account for the energy it took to raise the temperature of the block and table.

    An alternative is to say that the work done by the propelling force is equal to the gain in energy of the block and table top associated with the increase in their temperature. The work done by friction is zero.

    You can look at it either way, but you can't have it both ways. If you insist that the friction force does work, it is not possible to explain where the energy is going.
     
  18. Oct 9, 2018 #17
    so, simply put, through friction, the KE is transferred in to heat? just as the "heat " in an engine, is transferred into force that created the KE in the first place.
     
  19. Oct 9, 2018 #18

    Mister T

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    It's converted to internal energy, or what is sometimes called thermal energy.

    Heat is energy that's transferred because of a temperature difference. We don't have that here. We have an adiabatic process.
     
  20. Oct 9, 2018 #19

    jbriggs444

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    You can account for it just fine. You have an external energy input to the system. Your hand on the face of the block is an external force doing work. You have an internal force pair where the force pair is doing negative work. Since it is an internal interface, energy is conserved. However this force pair is sapping mechanical energy from the system and replacing it with thermal energy.

    If you want to be more complete in the analysis, you should drop the implicit assumption that the table top defines the lab frame and include all external forces. You have your hand on the face of the block and the floor anchoring the legs of the table. Those two external forces do an total amount of work that is positive and frame-invariant. The internal frictional force pair does an amount of work that is negative and frame-invariant. The result is that no net work is done, regardless of what inertial frame is adopted and that thermal energy is continuously incremented regardless of what inertial frame is adopted.
     
  21. Oct 9, 2018 #20
    Well ~ to say for sure, friction in the real world is always a positive force ; if it wasn't so! we could not be moving when walking .
     
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