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Work equation?

  1. Aug 3, 2004 #1
    [tex]\textcolor{red}{W=\vec{F}dcos \theta}[/tex]

    If the object is not moved by the force applied, I'm sure we can all agree that [itex]\textcolor{red}{W=0~Joules}[/tex], ergo no work was done. Does this also imply that there was no energy output?

    Scenario 1:
    A man applies [itex]\textcolor{red}{4N}[/itex] of force on an object. In the end, he moves the object a total of [itex]\textcolor{red}{2m}[/itex] horizontally. So from this we can say: [itex]\textcolor{red}{W=(4N)(2m)cos(0)}[/itex]. This gives us: [itex]\textcolor{red}{W=8~Joules}[/itex]. Agree? There was work done, and there was also energy output by the man in a total amount of [itex]\textcolor{red}{8~Joules}[/itex].

    Scenario 2:
    A man applies the same amount of force on another object. This object is much more massive than the first object. He eventually gets tired of trying to move the object, and can not go on. So, he accomplished to move the object [itex]\textcolor{red}{0m}[/itex] across a horizontal surface. So, from this we can say: [itex]\textcolor{red}{W=(4N)(0m)cos(0)}[/itex]. This gives us: [itex]\textcolor{red}{W=0~Joules}[/itex]. There was no work done. Does that mean there is no energy output since the amount of work done was [itex]\textcolor{red}{0~Joules}[/itex]? Obviously, this man used energy to TRY and push the object. He was tired, because he output all his energy trying to move the boulder, but the equation says otherwise.

    Scenario3:
    [tex]\textcolor{red}{\vec{F}=G\frac{M_1M_2}{r^2}}[/tex]

    The earth pulls on the moon with a certain amount of force. The moon pulls back with that same amount of force. This allows the moon to orbit the earth. Does the force that keeps the moon in orbit apply work?

    [tex]\textcolor{red}{G=6.67300x10^{-11}N-m^2/kg^2}[/tex]
    [tex]\textcolor{red}{M_1(Earth)=6x10^{24}kg}[/tex]
    [tex]\textcolor{red}{M_2(Moon)=7x10^{22}kg}[/tex]
    [tex]\textcolor{red}{r_{distance}=3.844x10^8m}[/tex]

    After calculating all that out, [itex]\textcolor{red}{\vec{F} \approx 18.9x10^{19}N}[/itex] That's a lot of force in the earth-moon system. It would seem logical to say that this force outputs energy inorder to keep the moon in orbit, right? If that were the case, where is this energy coming from? To get around this work is introduced into the situation. The moons orbit is not perfectly circular, but its eccentricity is 0.0549. That is neglible. So, if we plug and chug in the work equation we get [itex]\textcolor{red}{W=0~Joules}[/itex]. We know that is not true though, because energy is used to keep the moon in robit. The reason why most scientists believe that no work is done is because if energy were output, there would be no source of this energy to drain, ergo violating the laws of conservation of energy.

    There is obviously something wrong with this. No, this is not an attempt to try and go against physics. I just want answers.

    I want just a sipmle explanation. Not scrutiny from everyone.
     
  2. jcsd
  3. Aug 3, 2004 #2
    Your analogy of a man pushing a rock is not a good example. The man is a complex mechanism with every cell doing work just to stay in existance.

    When you get a good handle on it you will see that the moon staying in orbit does not require work except for the tides both on the moon and the earth. There's nothing wrong with the equations. We just have to understand them better.

    Keep on chuggin !!

    Vern

    Gravity Works
     
  4. Aug 3, 2004 #3

    russ_watters

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    Actually, the analogy works fine - what he's missing is the difference between input and output. The calories burned by the man are energy input, the work done on the rock is work output.

    Sounds strange, doesn't it - the earth applies a force causing the moon to be in constant acceleration - yet no work is done. But thats the way it sometimes works.
     
  5. Aug 3, 2004 #4
    I know the difference between input and output, and those are adressed in my problem.

    That's simply not enough. There is something wrong with the work equation, or gravity is flawed. The work equation is what is used to cover up the blunder that there is no source of input energy for the earth on the moon. Seeing that the work equation get's 0 work, there is no need for a source of energy with no output energy. It is well known that the earth is using energy to keep the moon in place, but where is this energy coming from?

    Vern, did you make up the photon theory? If so, I'll have to hand it to you. You are questioning way things work; that's always good. Whoever made the site, I like it. Vern, my analogy is plausible.
     
  6. Aug 3, 2004 #5

    chroot

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    1) People are complex. When you push on something, your individual muscle fibers contract and relax over and over again. This expends energy and tires you out even when you don't move anything.

    2) Pushing a box horizontally does no work against gravity. On a frictionless plane, it requires zero energy to push a box from one place to another.

    3) Gravity does do work on orbiting bodies. Imagine a 100 kg satellite in an eccentric orbit with an apogee of 10,000 miles and a perigee of only 1,000 miles. That means that in going from apogee to perigee, the satellite loses 9,000 miles of altitude. Gravity does work on the satellite as it loses those 9,000 miles of altitude, giving it a little over 14 billion joules of kinetic energy. (Making the simplification that g does not change over that altitude.) As the satellite passes its perigee, it begins gaining altitude again, losing speed and doing work against gravity.

    The critical thing you must realize is that the total energy is conserved. During the descent from apogee to perigee, gravity is doing work on the satellite. During the ascent from perigee to apogee, the satellite is doing the same amount of work against gravity.

    If you roll a ball up the side of a bowl, it'll roll back down again, and then up the other side. And then down and back up the other side. And then down and back up the other side. All with no energy input from you other than the initial amount to get the ball rolling. In the absence of friction, it would go on rolling up and down forever. A satellite in orbit around the Earth experiences negligible friction, and thus orbits forever. All with no energy input from you other than the initial amount to get it going.

    - Warren
     
  7. Aug 3, 2004 #6

    chroot

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    No, you simply don't understand it well.
    It uses no energy at all.

    - Warren
     
  8. Aug 3, 2004 #7
    How can it do work? The output energy of earth doing work is through the forces. These forces arise from attraction. So, where does this energy come from?
     
  9. Aug 3, 2004 #8

    chroot

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    You launched the satellite didn't you? With a big ol' rocket full of explosives? That's where the energy came from.

    - Warren
     
  10. Aug 3, 2004 #9
    Earth-moon system only has forces through attraction. There is no initial input of energy, yet the earth and moon are able to exert forces. There is no source of this energy input for an output of energy in the form of work.

    When I proposed the experiment of the man pushing boxes, I forgot to add that the surface was not frictionless. Yes, I know how muscles work. I didn't take BioII AP for nothing. This energy that is loss through muscle contraction is in the form of input energy. The only way to get an output is to move something. If nothing is moved, there is no output but only input. This is what the work equation says.
     
  11. Aug 3, 2004 #10

    chroot

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    Of course there was. The solar system didn't just come from nothing, did it? No, indeed, it formed from the collapse of a large cloud of gas and dust. The collapse provided lots of energy.
    He still didn't do any work against gravity. He just did work against friction. The equations you gave were wrong, as they used an angle of 0 degrees between the normal force and the applied force. (It should have been 90 degrees.) You didn't include a friction term in the equation, even though you now claim to have been talking about friction. :rolleyes:
    No. You can also produce heat (you'll get sweaty), sound, deformations in your tennis shoes, etc. The energy you expend pushing against a wall goes into all of these forms.

    - Warren
     
  12. Aug 3, 2004 #11
    Sorry, Warren.

    Scenario 1:
    [tex]W=4N*2m*cos(90)=-3.58458893~Joules[/tex]

    Scenario 2:
    [tex]W=4N*0m*cos(90)=0~Joules[/tex]

    Scenario 1 yields a negative energy output. While scenario 2 still yields no energy output.

    Warren, all of the energy you input comes from your body as one system. When you input ATP to apply calcium to the actin-myocin bridge to yield output, that is a system of it's own. The output of this system is on the object which the force is applied. Whether this energy is going into the wall or to your shoes, there is no movement of the object. There is no ouput but input is present. If there is no output, there should be no input. The work equation accounts for the total amount of energy you spend regardless of where it goes into moving that object. In Scenario 1, you moved the object, although the total amount of input energy did not go directly to moving that object.
     
  13. Aug 3, 2004 #12

    chroot

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    You seem quite confused. The cosine of 90 degrees is zero. Perhaps you should try doing your math homework.

    There is always output. It may be heat, sound, etc. I've already said this. Do you not agree that heat is a form of energy?

    - Warren
     
  14. Aug 3, 2004 #13

    Chronos

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    This is no different than asking where a permanent magnet get its energy from. It exerts a measurable force on any nearby magnetic object. But, it doesnt wear out, it doesn't get hot and it is way, way more powerful than gravity.
     
  15. Aug 3, 2004 #14
    Whoops, hehe. Last time I rely on google for calculations.

    I agree that heat is a form of energy. If no work is done, there is no output energy going by the equation. It is obvious that there is energy that is output as you said in the form of heat. This is neglected going by the equation. If you are putting in a total of 4N, you should output equivalent to 4N regardless if you move the object or not. The work equation yields no output in any form of energy. The work equation is not valid for someone who does not move an object with input force.
     
  16. Aug 3, 2004 #15

    chroot

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    http://www.google.com/search?num=30&hl=en&lr=&ie=UTF-8&safe=off&q=cosine(90+degrees)&btnG=Search
    Your equation is not complete. Duh.
    The Newton is a unit of force, not work (or energy). There is no law of conservation of force. :rofl:
    The proper equation to use is simply

    [tex]\Delta E = 0[/tex]

    The total energy of the system is conserved.

    - Warren
     
  17. Aug 3, 2004 #16
    This is where you go wrong. Assuming there is no friction between the massive object and the ground the massive object does move. Only that if its massive enough say like if you push on an asteriod, you wouldn't see any movement because it moved to little for you to see.

    If there is friction then the energy is transfered against the molecular bonds in the atoms of the ground.
     
  18. Aug 3, 2004 #17
    Ehhh, Warren, I'll avoid radians next time. cos(90) does not work alone. :redface:

    That is not my equation. It is an equation that was first thought to help engineers. I know Newton is a unit of force. It is an input force that requires an energy to be existent. Notice I said "equivalent."

    No, energy is not conserved. You have an input which consist of no output going by the work equation.

    Entropy, I agree with you on both points. With the presence of friction, it transferred to the ground. This is the amount of work done, but is not yielded by the work equation, if an object is not moved. Thios transfer of friction to the ground is apart of you applying work to the object.
     
  19. Aug 3, 2004 #18

    chroot

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    :rofl: What does that even mean?
    As I've already said, your equation is not complete. You only have a term for the movement of an object against a force, but you've neglected to include terms for heat, sound, etc. In the case of a person pushing against a wall, those terms are non-zero, and cannot be neglected.

    This is not difficult to understand.

    - Warren
     
  20. Aug 3, 2004 #19
    To Ultra
    I often get mixed up in questions concerning rotation -- but i've learned that's usually because I mix frames of reference. If you stand on the earth surface and look at the moon it's clear that it is not rushing either toward or away from us , and yet you can show that gravity is at work by dropping an apple. So if gravity is attracting the moon how come it does not fall. Within that framework we say that a force ( centifugal outward) exists which balances gravity in such a way that no accelleration occurs and no distance change -- hence no work is done.
    You may say OK but where does this mythical force come from and why does it not appear to work for the apple ?
    To see this you have to change your perspective viewpoint to a stationary one
    outside of the rotating system . In this frame it is natural for massive bodies to travel in 'Straight lines' unless force is applied so that the moon in going around the earth must be feeling a force ( gravity ) to deviate it and is in fact undergoing accelleratations all the time causing it to undergo a back and forth movement which as Chroot says is a motion which causes the exchange of Kinetic and potential energy
    even if it is perfectly circular.
    IT ends up the same way that is the total energy is constant .
    If you picture this motion from the perspective of the suns surface you will see that the moon during one month the moon undergoes a cycle of approach to the sun follwed by gaining distance , it is sinusouidal where the accelleration is greatest at nearest and furthest approach and least at maximum speed ( wrt the sun).
    Hope this helps Ray.
     
  21. Aug 3, 2004 #20
    i have come across reports that say the moon is getting farther away from the earth by like .1 cm.. or something measurable like that.. wouldn't this mean that the centrifugal force was overtaking gravity?
     
    Last edited: Aug 3, 2004
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