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Suppose I am able to lift a body of mass $m$ up using a constant force $F=mg$ so that net acceleration is zero on it. I want to find the potential energy and the total energy of the body when I have made the body travel a height $h$.

Easy. Gravitational Potential Energy $= mgh$. Kinetic Energy $= 1/2 m v^2$. Add them up and I get total energy.

Now. Seeing this a bit differently. Work done by Force $= mgh$. Work done by Gravity $= - mgh$. Net Work Done $= mgh -mgh = 0$ N. But potential energy $= mgh$.

How is this potential energy achieved when the net work done is zero? This is my doubt. I know I'm missing some basic concept here but please help.

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  • $\begingroup$ Please see our guide on writing good titles. $\endgroup$
    – user10851
    Apr 12, 2016 at 2:44

3 Answers 3

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You are not defining your systems clearly enough.

If the system is the mass and the Earth then in the situation that you have described the external force doing work on that system increases the potential energy of the mass-Earth system.

Now look at the system which is the mass alone. Then there are two forces acting on the mass. The force as described above and the force due to the gravitational field (weight) which are equal in magnitude but opposite in direction. So the net force on the mass is zero and thus there is no change in the kinetic energy of the mass.

Perhaps you are forgetting that a single mass by itself cannot have gravitational potential energy? It is often stated that a mass has gravitational potential energy $=mg\Delta h$ where it should be stated or understood that the mass-Earth system has the potential energy. The confusion might well be compounded by the fact that when a mass is dropped it is seen to move faster and so gains kinetic energy. What is forgotten is that, although not measurable, the Earth also accelerates upwards towards the mass and gains kinetic kinetic energy. So it is both the mass and the Earth which are losing potential energy.

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  • $\begingroup$ So there is no work done by gravity? $\endgroup$
    – Quark2
    Apr 11, 2016 at 6:52
  • $\begingroup$ If you are considering the mass-Earth system the force on the Earth due to the gravitational attraction of the mass and the force on the mass due to the gravitational attraction of the Earth are equal and opposite internal forces. $\endgroup$
    – Farcher
    Apr 11, 2016 at 6:58
  • $\begingroup$ So, I committed a mistake in ignoring the force on earth due to the mass which cancels out the gravitational force since I was confused between various frames. Am I correct? $\endgroup$
    – Quark2
    Apr 11, 2016 at 7:08
  • $\begingroup$ You have to be careful with the words "cancels out" because the attractive forces act on different bodies. Yes, the net work done is zero. $\endgroup$
    – Farcher
    Apr 11, 2016 at 7:15
  • $\begingroup$ Newton's third law tells you that is so. $\endgroup$
    – Farcher
    Apr 11, 2016 at 7:24
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By work-KE theorem, $$\Delta \text{KE} = \text{Work done by NET force}$$ $$=\text{Work done by} F_1+\text{Work done by} F_2+\cdots$$

Now if a certain force $F_i$ is conservative, you have the choice of defining its corresponding PE so that

$$\text{Work done by}F_i=-\Delta \text{PE}_i$$

and MOVE it to the left hand side so that you have

$$\Delta\text{KE}+\Delta\text{PE}_i= W_1+W_2+\cdots+W_{i-1}+W_{i+1}+\cdots$$

Notice that in doing this $W_i$ disappears from the RHS.

In conclusion, you either consider it as work done or (if the force is conservative), consider it as PE, BUT NOT BOTH, as what you have done in your argument.

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Potential energy is not energy. If you, for example, place an object of mass m on your outstretched palm and raise it through a distance h, so that it is initially stationary and ends up being stationary, then the total work done on the object by the weight force acting on it is -mgh and the total work done on it by the normal force acting on it due to its interaction with your palm is mgh. The total work is 0 and the change in energy is 0.

Potential energy is a concept used in some kinds of analyses because it can make them easier to perform and/or understand. It is not energy.

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  • $\begingroup$ It has units of energy and is called "energy" ("kinetic energy" and "potential energy"). Walks like a duck and quacks like a duck... $\endgroup$
    – user12029
    Apr 11, 2016 at 18:00
  • $\begingroup$ In the same way that the term "Potential winnings" describes things with the same units as winnings and contains the word "winnings", but you can't use them to buy anything. $\endgroup$
    – user113683
    Apr 12, 2016 at 13:00
  • $\begingroup$ Just to clarify, you can treat potential energy as energy in the question posed above, but then you have to neglect the work done on the object by the weight force acting on it. In problems such as this, that causes unnecessary confusion for those who question what they are doing (that's why students often ask questions like the one above), and mistakes when people don't question. $\endgroup$
    – user113683
    Apr 12, 2016 at 13:17

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