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Calculate the Energy or work

  1. Jun 16, 2005 #1
    Hello,

    I need to calculate the Energy or work involved in this situation :
    An object of mass M moves from point A to B at a constant speed.
    (initial speed is zero).
    So I only know the mass (M),distance AB,initial and final speed (0,V) and the time taken (T).
    What's the formula I should use to calculate the Energy used by the object?

    bye bye.
     
  2. jcsd
  3. Jun 17, 2005 #2
    ahem. constant speed? how can it move at a constant speed if it's initial velocity is zero? if you mean constant acceleration then it's simple. you know the distance, final velocity and time taken. calculate the acceleration and then calculate the work done from there.
     
  4. Jun 19, 2005 #3
    Ops,I meant : constant speed . The initial velocity is not zero,it is constant. Sorry.
    Now, knowing that there is no acceleration,how do I calculate it ?!?
     
  5. Jun 19, 2005 #4
    Does [itex]W = Fd[/itex] apply in this situation?
     
  6. Jun 19, 2005 #5
    There is no work done on the object.
    It's energy is purely kinetic.
     
  7. Jun 20, 2005 #6
    yeah, W=Fd is applicable. Here the body moves with constant speed, so the net force on it is zero, hence the net work done is zero.
     
  8. Jun 21, 2005 #7
    I suppose there should be kinetic energy,but the velocity doesn't change so,the total energy is zero,but it doesn't make any sense because there must be energy "somewhere". Any hints ?!
    The problem is simple but I just can't solve it.
     
  9. Jun 21, 2005 #8
    The energy is purely kinetic energy.
     
  10. Jun 21, 2005 #9
    One of two things can happen here horazio.

    If there is no change in potential energy along the direction of motion, (Like moving side ways along the earth), then the work done is zero. This is because there is no change in potential energy as you move side ways, so it takes no effort to keep it moving in constant speed. If you call the horziontal direction the x direction, and the verticel the y direction, then if you throw ANYTHING in the horiztonal direction, it will keep that same speed.
    The other case is if you throw it up or down. This is different because you are working against gravity. Here the potential energy changes. So if you want to move with a constant verticle speed, you must apply a constant force that opposes gravity for some distance d. So even though you have "constant speed", the energy is being used up via potential energy as you raise the object. But this is not limited to gravity, it can be any type of field.

    The only thing here is to pay close attention to HOW your moving. If your moving against a field, energy is changing. If you are moving perpendicular to it, (like moving horzinotally), then your energy is not changing.
     
  11. Jun 21, 2005 #10

    Work is done by any kind of force, not just a field.
     
  12. Jun 21, 2005 #11
    Yes thats true, but I was trying to show him why a energy does increase, because in his example it is not kinetic energy that is increased as you stated, it has constant velocity. The potential energy is what is increasing for his example to work, actually friction could be another example too.
     
    Last edited: Jun 21, 2005
  13. Jun 21, 2005 #12
    I said there was no increase in energy. The total energy in the object is kinetic. You are making assumptions about PE and friction whereas he didnt mention anything about them.
     
  14. Jun 21, 2005 #13
    My mistake sorry whozum. Think about the problem though, you CANNOT have a force and CONSTANT velocity without a counter force, i.e friction OR gravity, can you? :-) Newtons First law demands it, the sum of all forces must be zero!
     
  15. Jun 21, 2005 #14
    No, you cant, but that has nothing to do with this problem.

    There were no forces mentioned.
     
  16. Jun 21, 2005 #15
    Yes, thats why I said there is one of two cases, where the work could be some value, or it could be zero, depending on how the equilibrium is maintained. One case requires a force, the other does not.
     
  17. Jun 22, 2005 #16
    Very simple.. the formula u should use is [tex] 1/2 m v^2 [/tex] to get the energy used up by the object.. not forgetting that this energy is used to overcome friction on the mass.
     
  18. Jun 22, 2005 #17
    yes.. there is a force, but it is used to overcome friction, therefore the sum of all the forces will be equal 0.
     
  19. Jun 22, 2005 #18
    Can someone please tell me where the OP mentioned anything about forces or friction..? I must have gone insane. This is such a basic problem and people are pulling things out of their ass.
     
  20. Jun 22, 2005 #19
    When there is no force, there will not be work done and when there is no work done, there will be no energy used..
     
  21. Jun 22, 2005 #20

    Doc Al

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    Staff: Mentor

    This question seems poorly phrased to me. What are you trying to calculate? The total work done on the object? If so, assuming we can treat the object as a particle and ignore changes in internal energy, then the total work done on the particle (by whatever forces may be acting on it) is zero since the kinetic energy doesn't change. (This is what whozum was saying, I believe.) Is this the question?

    As far as what is the "energy used by the object"? I don't know what that means.
     
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