Consider a particle on the ground. This particle is raised by a force of magnitude $mg$ to a height $h$ above the ground. At this point, the work done on the particle by the force is $mgh$, which is equal to the potential energy of the particle. But, during this period, the force of gravity also acts on the particle and is displaced by $h$, and so does a work of $-mgh$ on the particle. Shouldn't the two cancel and no net work should be done on the particle?

If they don't cancel, then where did the energy that came from the work done by the force of gravity go?

  • $\begingroup$ Note that work is a scalar quantity. Will vector addition laws apply? :) $\endgroup$ – mikhailcazi Feb 14 '14 at 11:51
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    $\begingroup$ Note that the definition of work $W=\int_{\gamma} {\bf F}\cdot\mathrm{d}{\bf r}$ depends on which force ${\bf F}$ it refers to, e.g. gravitational force, friction force, net force, etc. Gravitational potential energy by definition only refers to the work of the gravitational force, and not any other force. $\endgroup$ – Qmechanic Feb 14 '14 at 11:57
  • $\begingroup$ @Qmechanic: Could you elaborate? $\endgroup$ – Gerard Feb 14 '14 at 12:22
  • $\begingroup$ @mikhailcazi: I don't believe I used vector addition. $\endgroup$ – Gerard Feb 14 '14 at 12:23
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    $\begingroup$ The change in energy is calculated by doing that; it's not really net work. If you remember the work-energy-theorem: $W_{conservative} + W_{non-conservative} + W_{other} = \Delta K.E.$ $\endgroup$ – mikhailcazi Feb 14 '14 at 12:57

Think about the work-kinetic energy theorem, which states that the net work done on an object is equal to its change in kinetic energy: $$W_{net}=\Delta\mathrm{KE}.$$

You are right that when lifting an object of mass $m$ by a height $h,$ in a uniform gravitational field, the work you do is $W_{you}=mgh$ (assuming, as you said, that you're applying a force of $mg$), and for that same displacement, the work done by gravity is $W_{grav}=-mgh.$ The fact that these two cancel out ($W_{net}=W_{you}+W_{grav}=0$) means that the kinetic energy of the object after being lifted is 0. So the work done by gravity went to sucking energy out of the object that you were adding, thereby converting it to gravitational potential energy. (If it did not get sucked out, then the object would gain kinetic energy--just imagine a case where you apply the same force to the object as you did when you lifted it, but this time there is no gravitational field. Then, since there will be a net force on the object in that case (or, net work will be done (by you)), the object's KE will increase.) Meanwhile, the change in gravitational potential energy of the object is $\Delta U = -W_{grav}$.

So, to summarize, you do positive work on the object by lifting it. Normally, that work would cause the KE of the object to increase. In this case, though, we have gravity doing negative work simultaneously, so the work you're doing is getting converted to potential energy by gravity. No net work is done, so the object has no KE after being lifted.

  • $\begingroup$ Excellent answer. Just to clarify, What is the exact definition of a 'field' (magnetic fields, gravitational fields, etc.)? If we push an object on a rough surface, the frictional force does negative work, sucking kinetic energy out. But it also produces heat, which gravity does not. I'm guessing this is because the object gains potential energy. But when does this potential energy exactly come into existence? $\endgroup$ – Gerard Feb 14 '14 at 12:54
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    $\begingroup$ A field (in physics) is a quantity with a value at each point in space. You can have a vector field, a temperature (scalar) field, etc. The connection between a field and a potential energy is that fields associated with conservative forces may be written as the gradient of some physical quantity (e.g. grav. potential), and if you do a line integral in these fields (e.g. calculate work done), the result depends only on the start and endpoints. Gravitational fields are conservative, while magnetic fields are not, for instance. Work done by friction is path-dependent, so it's not conservative. $\endgroup$ – Mike Bell Feb 14 '14 at 13:18
  • $\begingroup$ So magnetic fields are not conservative because they can't be written as a gradient, right? They have different shapes altogether. $\endgroup$ – Gerard Feb 14 '14 at 13:22
  • $\begingroup$ You can define a potential energy for conservative forces/fields, but not for non-conservative forces like friction. So in the case of a (conservative) gravitational field, the potential energy comes into existence by the fact that gravity is a conservative force/vector field (so the potential energy is defined). Whenever the conservative field does +/- work on an object, the potential energy of that object changes. When a non-conservative force like friction does work, that energy goes to heat, as you said in the comment above. $\endgroup$ – Mike Bell Feb 14 '14 at 13:24
  • $\begingroup$ That's right, magnetic fields can't be written as a gradient. They can be written as a curl instead. $\endgroup$ – Mike Bell Feb 14 '14 at 13:27

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