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Ok, so I am in the 10th grade, and Physics is my life, and so I always try to enquire things. We were studying the chapter of Work and Energy during our Physics lecture, and when our lecturer described the derivation of formula to calculate the Potential Energy (U) = mgh, something struck me. The derivation goes in this way :- Gravitational Potential Energy, is measured in terms of work done in bringing a body from the ground level to a certain height against the forces of gravity, the work done then gets stored in the form of Potential Energy in the body. Therefore, in our textbook it was given, Work = Force x Displacement The least force required to lift the body from the ground will be mg i.e. its mass x acceleration due to gravity(this part is bothering me). Therfore the Force will be mg, while the displacement done by the force will be the height at which it was raised against the forces of gravity that is h. Therefore :-

Total Work Done in lifting the body = mgh (force x displacement) Therefore, Potential Energy = mgh.

But isnt it true that the least force required to lift the body should be greater than mg. Since the force of gravity acting on that body too will be mg but in opposite directions and these two forces should cancel each other and the net force will be zero, therefore displacement will be zero and the body will remain in equilibrium. I dont know if this is stupid but can anyone pls answer.

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Since the net force acting is zero the , from Newtons 2nd law the momentum is conserved and hence the object moves with a constant velocity but that's only till you keep on applying force for its entire journey . Once the applied force is removed the velocity decreases in the direction of motion since acceleration is in the opposite direction and the energy that you apply to the object is subtracted from the gravitational potential energy of the earth-object system , which gives you the net potential energy .

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  • $\begingroup$ But dont you think, we would require a force greater than mg, to break the equilibrium, how can the momentum be conserved, it is like taking a body and stretching it from two opposite sides with an equal force, the body will have no displacement and hence no work will be done. $\endgroup$ Jun 24 '17 at 5:29
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This is a good question which is really asking as to how you can move a mass $m$ from position $A$ to another position $B$, a distance $h$ above position $A$, whilst at the same time doing no net work on the mass which has a downward force acting on it due to the gravitational attraction of the Earth - the weight of the mass $mg$.

Apply an upward force magnitude $F_{\rm external}$ on the mass equal in magnitude to the weight of the mass and start the mass off with an upward velocity $v$ at position $A$.
The mass will move up with a constant velocity when moving from position $A$ to position $B$ because no net work is being done on it (net force = $F_{\rm external}$ - mg =0) and the kinetic energy of the mass does not change.
You can of course have any value you like for the magnitude of the velocity.

With the mass starting from rest, apply a force $F_{\rm external}$ on the mass which is greater than the weight of the mass $f = F_{\rm external} -mg$ and after the mass has moved a distance $z$ reduce $F_{\rm external}$ so that it is equal in magnitude to the weight of the mass.
The work done on the mass $fz$ will equal the gain in kinetic energy of the mass. This will give the mass an upward velocity and kinetic energy which will stay constant as long as $F_{\rm external} =mg$.
When the mass is a distance $z$ from position $B$ reduce $F_{\rm external}$ so that the net force on the mass is $f$ downwards. The work done on the mass will now be $-fz$ and will equal the loss in kinetic energy of the mass.
The mass will stop at position $B$ because the net work done on the mass is zero $(fx-fx=0)$.
From this you can generalise and say that all you need to do is do an (infinitesimal) amount of work on the mass to get it moving when it starts off from position $A$ and then get that work "back" whilst the mass slows down and stops at position $B$.
When your teacher demonstrates the increase in potential energy by lifting a stationary mass on the floor and then placing it on a bench top this is what your teacher does.


Once you have convinced yourself that you can move the mass from position $A$ to position $B$ by doing no net work on the mass alone you can then apply the same ideas when you look at the mass-Earth system as the mass alone does not have gravitational potential energy, it is the mass and the Earth which has the gravitational potential energy.

So you either use the definition that the change in potential energy is equal to the work done by external forces in increasing the mass to Earth distance $=F_{\rm external} \times h = mgh$
or the change in gravitational potential energy is equal to minus the work done by the gravitational field in increasing the mass to Earth distance $= -(-mg \times h) = mgh$.

It is assumed that because the Earth is so massive compared with mass $m$ the movement of the Earth relative to the centre of mass of the mass-Earth system is very small compared to the movement of the mass and hence the work done by the force acting on the Earth can be neglected.

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  • $\begingroup$ Thanks a lot for your answer, I got it now, it means that we can move a body at a constant velocity from position A to position B by applying by applying a force equal to its weight, so the body will still remain in a Dynamic Equilibrium? $\endgroup$ Jun 24 '17 at 8:14
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There have been a couple of questions similar to this recently, which you should look at. As for your specific question, you have to accelerate the object upward in order to increase its potential energy, so you are inherently giving it kinetic energy by increasing its gravitational potential energy. In other words, the work you do goes into both kinetic and potential energy, but the work you did against gravity is only considered for its potential energy. The kinetic energy may turn into potential energy if it keeps moving once you stop applying a force, in which case the total work you did will all eventually go into potential energy at the point that it stops moving.

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This definition seems to be quite common in introductory textbooks. Probably is considered more intuitive but it usually creates confusion. But the actual definition of potential energy does not involve any external force but the one of the field for which we define potential energy. So the potential energy is the work done by the field (in this case gravity) when moving the body from the reference point to the actual position, with a minus sign. In your case, displacement from the surface to a height h, the force (gravity) is pointing downwards and the displacement is upwards. So the work is negative, W=-mgh and the potential energy is PE=-W= mgh. It does not matter how you move the body between the two points, the work done by gravity is the same. This is why we can associate PE with the gravitational field to start with.

Of course, if during this motion you act with a force exactly equal with mg but oriented upwards, the work of this force will be also mgh. But this has actually nothing to do with the gravitational potential energy. They just happen to have the same value.

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  • $\begingroup$ Yes, the work done by gravity can be measured as negative, so it will be -mgh, while the work done in lifting the body will be +mgh, on adding these two in order to calculate total work done we will get it as zero? $\endgroup$ Jun 24 '17 at 5:30
  • $\begingroup$ One more time, potential energy of a field is related to the work done by that field. Only. All the other forces are irrelevant. This is the definition of potential energy. The total work (done by all forces, also called net work) is related to the variation of kinetic energy. See work-energy theorem. So if net work is zero then the KE variation is zero but the PE of gravity I increases by mgh. $\endgroup$
    – nasu
    Jun 24 '17 at 13:16
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No @Ajinkya Naik in order to move a body you'll need to apply a force which is greater than or equal to the force holding it ( i.e mg ) . For example when an object is stationary on a surface having friction....you'll have to apply a force which is greater than or equal to the static friction holding it , once the force applied becomes equal to the value of static friction the object starts moving and after it starts moving if force is applied in proportion to the frictional force such that the net force is zero than the momentum will remain conserved for different instant of times . The case of an object on the earths surface is exactly the same only here the resisting force is gravity not friction . You should know this in order to derive the equation for potential energy we assume that the force is applied equal to the gravitational force acting such that the net force is zero hence the momentum is conserved but the displacement will not be zero since the object has moved . Hope this helps!

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  • $\begingroup$ yes, it helped me a little but it has still left me baffled, can you state the reason, how can a body be displaced if we are applying a force that is equal and opposite to its resisting force. I mean in the absence of an external force, even a slightest push can displace the body, but how can two equal and opposite forces cause motion? $\endgroup$ Jun 24 '17 at 5:57
  • $\begingroup$ The force applied must be very very slightly greater than the resisting force. $\endgroup$ Jun 24 '17 at 5:57
  • $\begingroup$ Yes @Ajinkya Naik that's true....we assume it's an approx equilibrium and for mathematical reasons it's a fairly good approximation. $\endgroup$
    – Munj Patel
    Jun 24 '17 at 6:24
  • $\begingroup$ yes, that is what I wanted to hear, it has a very little mathematical error. $\endgroup$ Jun 24 '17 at 7:08
  • $\begingroup$ All these considerations about the force being equal or slightly larger and so on are not relevant. The potential energy of gravity is defined in terms of the work done by the force of gravity. It does not matter if there is or not another force and if this force is equal to or ten times larger than the force of gravity. The change in PE is the same, as it is determined only by the work done by gravity. The work of the other force(s) may be equal in magnitude with the work done by gravity but this is just a coincidence not a necessity. $\endgroup$
    – nasu
    Jul 1 '17 at 21:18

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