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If the net work is zero. Why is the object still moving?

  1. Apr 28, 2013 #1

    I have a ball, and I'm lifting this ball upward with +work, and gravity is doing -work, Net Work= zero.
    But how is it, that I can still move the ball.
    When read that explanation I think of the ball or any object not moving at all :S

    So how can I figure this puzzle out :O?
  2. jcsd
  3. Apr 28, 2013 #2


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    The work you are doing in lifting the ball is converted into additional potential energy of the ball. This does not mean that the ball cannot move.
  4. Apr 28, 2013 #3


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    That's a very odd way to define "net work". Any process involving energy conversion -- which is the entire point of the field of thermodynamics -- utilizes a conservation of energy statement to equate the input and output work, either in the form of Win-Wout=0 or Win=Wout.

    But, I suppose if we're being pedantic anyway, there is an easy way out of this one. The rise in height represents potential energy, not work. So in this case you really have Win-Eout=0.
  5. Apr 29, 2013 #4
    I was watching a lecture in Physics by UCBerkley and there was an example of the total net work done on a "chalk", where the professor was lifting it above the ground doing +Work, and gravity is doing -Work on the chalk thus the Net Work = 0.

    So I was a bit confused.
  6. Apr 29, 2013 #5
    Ok, the work I'm doing on the ball is being converted to PE, that means their should be no motion :S
    Since the motion is caused from the work I put in... That ball should stop, and if I release it all the work I put in is converted from PE > KE + losses due to resistance.
  7. Apr 30, 2013 #6


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    Not sure what you mean here. The net work done by all forces acting on an object is equal to the kinetic energy change of that object. If the net work is zero, that means there is no kinetic energy change. That means that once the object being lifted is in motion, it moves up at constant speed.
    You can look at another example without PE, like pushing a block along a rough level surface. If the pushing force is 10 N and the value of the opposing kinetic friction force is 10 N, then there is no net work being done, but for sure, the block is moving. At constant speed.
  8. Apr 30, 2013 #7
    I think net work will be zero as according to law of conservation of energy.
  9. May 1, 2013 #8
    Since the force's cancel out.
    And that block has no "change" in KE, I assume it still has KE but not changing?
    I'm not used to "net work" I never considered the -Work done on a object as +Work is applied.
  10. May 1, 2013 #9

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    That's correct, assuming the piece of chalk is at rest initially and at rest after being lifted to some height.

    Suppose the piece of chalk starts at rest, the professor lifts it upwards a bit, and then stops. The initial and final kinetic energy of the piece of chalk with respect to the ground are both zero. This means there's no net change in kinetic energy. Per the work-energy theorem, the net work done on the piece of chalk is thus zero.

    Compare that to the situation where the piece of chalk is still being lifted upwards. The initial kinetic energy is still zero, but the kinetic energy at the point in time of interest (chalk moving upward) is non-zero. Now we do have a change in kinetic energy, requiring a non-zero net work between the halk moving state and chalk at rest state.
  11. May 1, 2013 #10


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    yes, while in motion at constant velocity, the pulling force on the chalk
    upward is balanced by the weight of the chalk downward. Or in the case of the block being pulled on a level surface at constant velocity,
    the pulling force to the right is balanced by the friction force to the left. No net force means no net work.
    Correct. As long as it is moving at constant velocity, its initial and final KE are the same, so there is no change during the motion.
    Net work is sometimes called total work. Work can be positive or negative. The pull force on the chalk moving up at constant speed does positive work, mgh. The gravity force does negative work, - mgh. The total work is 0. Or in the second example, the pull force does positive work , fd. The friction force does negative work, - fd. The total work is 0.
  12. Nov 19, 2014 #11
    I am sorry to bring up an old thread, but this question is bothering me. So, what is the total energy of the system? Is it only the potential energy of gravity? If so, why isn't the energy done by the person lifting the chalk up considered?
  13. Nov 19, 2014 #12


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    Assuming "the system" is the piece of chalk, the total energy is its kinetic energy (##\frac{1}{2}mv^2##) plus its gravitational potential energy (mgh)

    The work done by the person lifting the chalk goes into changing its gravitational potential energy (by changing its h), or into changing its kinetic energy (by changing its v), or some of both. $$KE_{initial} + PE_{initial} + W = KE_{final} + PE_{final}\\W = (KE_{final} - KE_{initial}) + (PE_{final} - PE_{initial})\\W=\Delta KE + \Delta PE$$
  14. Nov 19, 2014 #13
    Basically, the constant KE will be converted to PE, right?
  15. Nov 20, 2014 #14


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    Energy can neither be created nor destroyed, just transformed into different types.

    Suppose the chalk starts from rest and is lifted up a bit to point B, at which point it continues to be lifted up at constant speed to point C, and then finally stopped at point D. For simplicity, consider the motion between B and C only, where KE remains constant , but the PE changes. So since KE doesn't change, then how can it be transformed to PE?

    The change in PE is due to the external work done on the system (chalk) by the person doing the lifting.

    Looking again at the conservation of energy equation

    [itex]W_{nc} =\Delta KE+\Delta PE[/itex],
    where [itex]W_{nc}[/itex] is the work done by non-conservative forces (work done by the lifter in this case).

    Since there is no KE change here, then the above equation reduces to

    [itex]W_{nc} =\Delta PE[/itex]. So it is the work done by the lifter that changes the PE of the chalk.

    Rearranging the above equation,

    [itex] \Delta PE - W_{nc} = 0[/itex], or since total energy of an isolated system can never change per conservation of energy laws, then

    [itex] \Delta PE + \Delta E_{other} = 0[/itex], where the system is now defined as the lifter-chalk system and [itex] \Delta E_{other} [/itex] represents the change in forms energy other than PE and KE , that is, chemical/heat energy of the lifter ([itex]W_{nc} = -\Delta E_{other}[/itex]). That is perhaps the 'missing energy' you are wondering about?
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