1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

A doubt regarding work and energy

  1. Dec 14, 2011 #1
    I was just confused now. Is there work being done if I lift a bucket of water straight up?
    I know there is something about "work being done BY the bucket" and "work being done ON the bucket".

    Since I'm not moving horizontally does it mean that no work is done on the bucket?
    But I'm still doing work because I would get warm etc ??

    I also read somewhere that even if I carried the bucket straight up and walked, it would be 0 work done since the horizontal force has nothing to do with the lifting part.. is this true?

    So in what cases would it be true that the bucket is doing water?

    I'm really confused..
     
  2. jcsd
  3. Dec 15, 2011 #2
    Do you recognize that work = force x distance?
    When you lift the bucket you exert a force and the bucket moves through a vertical distance. You do work ON the bucket. It gains PE.
    If the bucket is dropped work will be done BY the bucket (it could be connected by a rope to some sort of machine !!!)
    When the bucket moves horizontally there is no change in height so no work is involved.
    If there was friction of some sort then work would need to be done to move the bucket horizontally.
    Very brief explanation, I hope it helps
     
  4. Dec 15, 2011 #3
    Oh great! yes it helps! Thanks :D

    but in the example u said, I do work ON the bucket but the bucket doesn't do any work right?

    or am I understanding it wrong?
     
  5. Dec 15, 2011 #4
    When you lift it you are doing the work......ON the bucket.
    You are providing it with potential energy.
    Work will be done BY the bucket if it falls and gives up this potential energy.
    If the bucket is simply dropped the you could say that the bucket is doing work on itself !!...(that is not a very technical phrase) the PE of the bucket will be converted to KE of the bucket....it will speed up.....OK with that ?
     
  6. Dec 15, 2011 #5
    yes

    thanks !
     
  7. Dec 15, 2011 #6
    Also, when you walk with the bucket horizontally at a constant speed, you don't need to apply any force to the bucket to do so. So you are doing no work on the bucket. However, if you are accelerating, then you ARE applying a force to the bucket, and you do work equal to that force times the distance you walk.

    If there is air resistance and you move at constant speed, then the force required to move the bucket simply cancels out the force of the air resistance, and any work you do to the bucket is canceled out by negative (opposite) work that the air does to the bucket.

    Another way to look at this is by looking at the bucket's kinetic energy. Since a change in kinetic energy correlates to a change in speed, the bucket at constant speed has constant kinetic energy, while the accelerating bucket is gaining kinetic energy. And by the work-energy theorem, we know that work done on an object is equal to the change in that object's kinetic energy. So at constant speed no work is done, but net work IS done on an accelerating bucket.
     
    Last edited: Dec 15, 2011
  8. Dec 15, 2011 #7
    Thank you!!

    It is very clear now :)
     
  9. Dec 15, 2011 #8

    PhanthomJay

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    Clarification: Work is done by forces and not by bodies, and by definition, is equal to the component of the force in the direction of the displacement time the displacement, positive if the component force and displacement are in the same direction and negative if the component force and displacement are in opposite directions.

    If the bucket weighs 20 N and the man lifts it with a force of 20 N (therefore at constant velocity) through a height of 2 m, then the work done by the (force of) the man is 20(2) = 40 J, and the work done by the force of gravity (the bucket's weight) is -20(2) = -40 J. The total (net) work done by both forces is therefore 0, consistent with the work-energy theorem which states that the change in KE is equal to the total (net) work done, for in this case, there is no KE change at constant velocity.

    I hope this clarifies rather than muddies the water.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook