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Problems On Work-Energy Theorem

  1. May 3, 2006 #1
    i have a Qns.

    given that i have a 2kg mass on earth and i want to accelerate from 0 m/s to 10 m/s the work done should be :

    Workdone = 1/2(m)(V)^2 - 1/2(m)(v)^2
    = 1/2(2)(10)^2 -0
    = 100 J

    however, if i were to take my frame of reference from let say space i have to include the speed of the earth orbiting the sun. then the equation turns out like this :

    let the speed of earth orbiting sun be= 100 m/s

    Workdone = 1/2(2)(110)^2 - 1/2(2)(100)^2
    = 2100 J

    so the Qns is why is there such a difference just because i take a different frame of reference? shouldn't the workdone be the regardless of my frame of reference??

    Thanks.
     
  2. jcsd
  3. May 3, 2006 #2

    Doc Al

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    Since KE and displacement are both frame-dependent, one would expect that the "work" done on an object would also be frame-dependent.

    What you are probably thinking about is a situation like this: Imagine a car (an ideal car, of course) on Earth being accelerated from 0 to 10 m/s. That requires a certain amount of fuel to deliver the needed KE. (Make it a toy car with your numbers: you need an amount of fuel that will deliver 50J of KE.)

    If KE is frame dependent, does that mean the amount of fuel needed depends on the frame in which one views the car? I trust you realize that the answer is: Of course not! :smile:

    To properly track the changes in KE, you must apply the "work"-energy theorem to both the earth and the car. In the earth frame, the earth is stationary, so no "work" is done on it. But from the space frame, both earth and car are moving, thus force*distance must be calculated for both. Try it and see what the net change in KE must be.

    Let me know if this addresses your concern.

    (I put quotes around "work" since it's really psuedowork--net force*displacement of the cm--that we are talking about.)
     
  4. May 3, 2006 #3

    nrqed

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    But it seems to me that one could consider a force acting on th ecar without acting on Earth. Let's consider the following "gedanken" experiment: A block is pushed (by an external force) by a constant force along a frictionless surface. Then the point of the OP holds. I think that the solution is somethig else but I need to do a quick calculation.
     
  5. May 3, 2006 #4

    nrqed

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    Indeed, the work done by the force will depend on the frame. So the obvious question becomes whether the W-E theorem holds if we change frames! It does. Using (let's assume only one force acting on the system):

    change of kinetic energy = force times distance

    then one finds that indeed the change of kinetic energy is indeed frame dependent but the distance travelled by the object is also frame dependent and when both are effects are taken into account, the two sides of the equations will be equal in any frame (neglecting relativistic effects of course).

    Prrof: Consider the left side in a certain frame S in which the object has an initial velocity v_i and a final velocity v_f. Now let's go to another frame S' moving at a speed v relative to the first frame. The change of kinetic energy (the left hand side of the equation) will increase by
    [itex] {1 \over 2} m (v_f+v)^2 - {1\over 2} m (v_i +v)^2 = m v (v_f-v_i)[/itex]

    Now consider the right hand side. In both frames, the acceleration is the same and is given by [itex] a = { v_f - v_i \over t} [/itex] so the force is [itex] m { v_f - v_i \over t} [/itex]. Now the *additional* distance travelled by the object as seen from S' is [itex] v t[/itex]. So the increase of the work done by the force as seen in S' is the product of those last two expressions, namely [itex] m { v_f - v_i \over t} \times v t = m v (v_f - v_i) [/itex]. This shows that the two sides of the equation will indeed change in a different frame, but by the same amount so that the W-E theorem will still be valid.
     
  6. May 3, 2006 #5

    Doc Al

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    I didn't interpret the OP's question as being "Does the work-energy theorem hold if one switches frames?" (of course it does, since it's just the integration of Newton's 2nd law). But no harm done in showing that! :smile:

    I interpreted the question as "How come it seems to take more energy to increase an object's speed just because I switched frames?" Which of course, it doesn't. (At least in the sense of how much fuel is required, as in my example.)


    Sure, that's a perfectly fine example and a similar analysis applies. I imagine a rocket with constant thrust pushing on the car. In that case, one must consider the "work" done on the rocket's exhaust as well as the work done on the car. (I'm sure I've looked at this exact problem before--maybe I can search the archives.) The amount of fuel that the rocket expends is independent of frame (of course) and a proper accounting of the work done on everything will show this.

    Maybe the OP will chime in and pick the answer that fits his question.
     
  7. May 4, 2006 #6
    thx for all the help!
    however i still have some Qns that i still don't understand( maybe the ans is there just that i don't understand) :

    Q1. what is psuedowork?

    Q2. Why is there a difference in the amount of WorkDone to, say accelerate
    an object from 0 m/s to 10 m/s, in different frames of reference?

    Q3. And if there is a different amount of WorkDone in different frames of
    reference, then won't fuel required to move an object change in
    different frames of reference?

    since ~WorkDone = Energy required for change of velocity~
    ( is the assumption correct in the 1st place?) ?

    + i understand than fuel require to change velocity is constant
    regardless of frames of reference( or our cars won't work), but
    why? since work-done in different frames changes shouldn't energy
    required to move car also change +
     
  8. May 4, 2006 #7

    nrqed

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    Ah ok. Yes, I did interpret the question in a different way so I went in a different direction.

    You are making a very interesting point. Actually, this is something I never had thought about (so thanks to you and to the OP for the question): when we talk about the work done by a force on an object, we really should include the work done by the force on all objects, including what I would call the "environment".
    For example, someone pushes on a mass. But that person has to have a "grip" on something in order to not move (for example, the friction on the shows keeps the person from slipping as he/she pushes a cart). In that last example, in the frame of the ground, the person does work on the cart but no work on the ground since the ground is not moving. However, as seen from a different frame, the person does work on both the cart and the ground.

    This is quite interesting:tongue2:

    Thank you

    Patrick
     
  9. May 4, 2006 #8

    nrqed

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    ouch...unless you know about virtual displacements and advanced mechanics I would suggest not to worry about this for now:wink:
    well, from a purely mathematical point of view it is simply because the force acts over a different distance as seen in different frames. That's all there is to it

    No. Think of a man pushing a cart and accelerating (let's say that friction on th ecart is negligible). The man is the source of the force. He is burning calories to do that. The work the man does on the cart is different in different frames.

    Now consider how much energy he is using up. The energy he is burning up is the work he does on the cart minus the work that anything else does on him. Is there anything doing work on him? well, in order to push the cart he must be "pushing himself" against something else (otherwise he could not apply a force for a certain time, he would just give a kick and flies away in the opposite direction as the cart). He can push the cart if he standing on the ground and there is enough friction to keep him on the ground, so he is pushing himself against the ground.

    Now, if you are in the frame of the ground, the ground is not moving so the work done by the ground on the man is zero. So the energy he will be burning will be equal to the work he does on the cart, *in thatframe*.

    Now, consider the situation from a frame moving with respect to the ground. Then the man does a different amount of work. But in that frame, *the ground is doing work on the man*. The *additional* work the man is doing (compared to what he does as measured in the frame of the ground) is equal to the work done by the ground on him. So the net result is that the energy he is burning up which is

    work done by him on the cart - the work done by the ground on him

    is the same as in the frame of the ground! Therefore he is burning the same amount of energy as seen from any frame.

    Hope this is clear. Thanks to Doc Al for pointing this out.


    Patrick
     
  10. May 4, 2006 #9

    Doc Al

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    What is pseudowork?

    I think Patrick did a good job answering your questions, but I'll just add a comment.
    The "work" that appears in the "work"-energy theorem is what I call pseudowork (also called, less mysteriously, "center of mass" work). It comes from taking the force on an object and multiplying by the displacement of the object's center of mass. Pseudowork is distinguished from "real" work, such as appears in the conservation of energy formula (1st law of Thermodynamics: [itex]\Delta U = Q + W[/itex]).

    Let's use my example of the accelerating car. For the car to accelerate, the ground must exert a friction force on the car. Using the work-energy theorem, you can calculate the resulting KE of the car. But is real work done on the car by the ground? No! The friction is static friction, the point of contact of tire with road is stationary, no work is done. This should make some sense, since the ground is not supplying the energy to move the car, the engine and fuel is! From a conservation of energy viewpoint, chemical energy (in the fuel/air) has been transformed into mechanical energy (KE) of the car.

    This is a bit subtle; don't get hung up on this term. :wink:
     
  11. May 5, 2006 #10
    Thanks for the help : )
     
  12. Jun 6, 2006 #11
    Erh.... i have another Qns regarding this issue.

    i get the general idea u are telling me.......the extra energy calculated in work-energy theorem in different frames is the "work done by the ground on him".

    i try to calculated( or account) the extra energy.......but i dunnoe how. can someone show me the workings ( take your own variables).

    p.s. i know that this is a very old post and u may have forgotten about the Qns, but pls help thx. ><
     
  13. Jun 6, 2006 #12

    Doc Al

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    Perhaps you can understand the frame dependence of work and energy more simply by analyzing this problem: An "ideal" car (no internal friction, air resistance) uses an amount of fuel to accelerate from 0 mph to V mph. We know the energy needed is [itex]1/2 m v^2[/itex]. And to accelerate from V to 2V, it takes an additional 3 times that energy (the total energy is [itex]1/2 m (2v)^2 = 2 m v^2[/itex]). But what if you viewed things from a frame moving at speed V? In that frame the second burst of speed only takes the car from 0 to V? Does the amount of fuel needed depend on the frame that you view things from? :wink: (Answer: Of course not.)

    Let's analyze this from each frame using forces and distances, and applying the "work"-energy theorem. Assume that the ground exerts a force F on the car to accelerate it.

    From the ground frame:
    The ground exerts a force F on the car. During the first burst of speed, the car moves a distance D, thus [itex]FD = 1/2 m v^2[/itex]. During the second burst of speed (from V to 2V), the car moves a distance 3D, so the additional energy needed is [itex]F(3D) = 3/2 m v^2[/itex].

    From the moving frame:
    The ground exerts a force F on the car, the car exerts a force F on the ground, and the ground (earth) moves! During the first burst of speed, the car's speed goes from -V to 0 (it moves backwards), thus it moves a distance -D. The change in KE of the car is thus [itex]F(-D) = -1/2 m v^2[/itex]. But during that time the earth had a constant speed of -V and thus moved a distance -2D. Thus the change in the Earth's KE is [itex](-F)(-2D) = m v^2[/itex]. The total change in KE is [itex]1/2 m v^2[/itex], the same in both frames.

    For the second burst of speed, the car's speed goes from 0 to V, moving a distance D, so it's change in KE is [itex]FD = 1/2 m v^2[/itex]. The earth still has a constant speed of -V, and thus moved a distance -2D, so the change in the Earth's KE is once again [itex](-F)(-2D) = m v^2[/itex]. The total change in KE is [itex]3/2 m v^2[/itex], the same in both frames.

    Make sense?
     
  14. Jun 7, 2006 #13
    yup thx again
     
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