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In physics, the potential energy of an object is mgh. We say that the potential energy of an object is equal to the work done on it to make it reach that height. So the work done = mg h. Mg is the force applied too lift the object to the height h But when we apply the force (mg) on an object, the gravitational force which acts on the object in the opposite direction balances the forces and the object doesn't move at all since the net force acting on the object is zero. This means that to increase the potential energy of an object or raise its height we have to apply a force which is greater than mg so that the gravitational force cannot cancel out the force fully. This means that to increase the potential energy of an object or raise its height we have to apply a force which is greater than mg . So why isint the equation for Potential Energy as P.E =Fh where F is a force >mg.


marked as duplicate by John Rennie energy Mar 7 at 7:38

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    $\begingroup$ Please do not say “X is wrong” when you are just learning about a physics concept that has been around for 300 years. This tends to irritate physicists. Thanks! There is a simple explanation which I will leave to others to provde. I’m fairly sure that this question has been asked and answered multiple times on this site. $\endgroup$ – G. Smith Mar 7 at 5:23
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    $\begingroup$ Related/duplicate $\endgroup$ – Farcher Mar 7 at 5:59
  • $\begingroup$ I think downvoting this is unkind. This is a common confusion for students and providing a convincing answer isn't trivial. $\endgroup$ – John Rennie Mar 7 at 8:11
  • $\begingroup$ @JohnRennie I wouldn't say that. Downvoting is never unkind in the sense meant in the code of conduct. But you could try to convince people who are downvoting it to adjust the standards they use to decide whether to downvote, such that they would change their mind about this one. $\endgroup$ – David Z Mar 7 at 8:44

If $F>mg$, the object will >>accelerate<< upwards under the net upwards $F-mg>0$ force. So it reaches height $h$ with some upwards speed $v$. And then its energy at that point is the sum $mgh+\frac12mv^2$ of potential plus kinetic energy. So $F=mg$ is the way to get a potential energy contribution only.

To get it moving, just assume it starts at $h=0$ with initial speed $v>0$ upwards, and thus with total initial energy $E_0=\frac12mv^2$ (no potential energy at $h=0$). Then upwards force $F=mg$ means the net force is zero, so the object continues upwards with its same, unchanging speed $v$. When it reaches any height $h>0$, its total energy is $E(h)=\frac12mv^2+mgh$, same $v$ and $\frac12mv^2$ as initially. So the difference $\Delta E=E(h)-E_0=mgh$ is just the potential energy due to the applied force $F=mg$.


If you do believe the work done by gravity to fall the object is less than the work to get it up there, then where do you think the rest of the energy go? The atmosphere never disappears on it way down nor is anything else involved. If you're still in doubt, try taking the long way out. Start by claculating the force on the object at that height: $$ F=G\frac {M_e m}{(R_e + h)^2}$$ or for simplicity, the usual $$ F=mg$$ Now use the force to calculate the acceleration of the object as it falls: $$ F=ma$$ Calculate time of fall: $$ h=at^2$$ Then the velocity when it touches the ground (zero air resistance assumed): $$v=at$$ Then kinetic energy when it touches the ground: $$K=\frac 12 mv^2$$

Then your newly found energy should be equal to the potential energy ($mgh$) before the mass start falling according to the law of conservation of energy (THE MOST UNIVERSAL LAW IN THE UNIVERSE). How about that? Crunch some numbers and validate that yourself.

  • $\begingroup$ If you take the force to depend on $h$ as in your first equation, you don’t get $h=a t^2$ (or even $h=h_0-a t^2$). The $t^2$ behavior comes from assuming a constant force, not one that varies with height. $\endgroup$ – G. Smith Mar 7 at 6:43
  • $\begingroup$ Considering the magnitude of $R_e$, the variation is quite small. That I believe is why the gravitational acceleration constant of earth is set at 9.8, because it's not that it doesn't change with height, but averagely we can practically let that slip. If you were doing this, what modifications will you introduce then? $\endgroup$ – TechDroid Mar 7 at 7:10
  • $\begingroup$ Then why not just say $F=mg$? The exact equation is soluble. See en.wikipedia.org/wiki/… But of course near the surface everyone uses the constant-force approximation, $\endgroup$ – G. Smith Mar 7 at 7:15
  • $\begingroup$ I thought maybe providing the original fundamental formula might help the OP grasp the situation better. And why do we small reputation guys have to be corrected or learn from larger reputation guys by loosing some of ours. Why can't you guys help out without down voting. Now I intend to tweak my answer, but the thought of the down vote still sticking around regardless makes it not so much worth it net-wise. You guys should be nicer. $\endgroup$ – TechDroid Mar 7 at 7:20
  • $\begingroup$ The main problem with your answer is that it never concludes that the potential energy is $mgh$, so it really doesn’t answer the OP’s question in my opinion. $\endgroup$ – G. Smith Mar 7 at 7:22

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