Suppose, I have an object of mass $m$. By applying a force of $F=mg$, if I withdraw this object with constant velocity and place it at a height $h$ from ground level, then potential energy stored in it will be $mgh$. If I apply the force more than $mg$, what will be the amount of potential energy stored in it? If in this case (when I will apply force more than $mg$ to place it at height $h$ from ground level), the potential energy equals to $mgh$, how it is possible?

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    $\begingroup$ If you apply a force more than $mg$, then the object will have kinetic energy, too, won't it? $\endgroup$ – ACuriousMind Jan 12 '15 at 15:51

I think it's more useful to consider this problem in a back-to-front manner. Let's consider the energy of the initial and the final state, and see what force is required to transition between them.

Initial state: Assume you define the initial position of the object to be at $y=0$. If the object is stationary, it has no kinetic energy, and so the total energy is the potential energy, which is $E_\text{i}=mgy=0$.

Final state: The object is now at the height of $y=h$. As you stated, the potential energy of the stationary object is $E_\text{f}=mgh$. The total energy of the object is just this potential energy, because it's not moving.

To get from the initial state to this final state, you need to add $\Delta E=E_\text{f}-E_\text{i}=mgh$ units of energy to the object. You can accomplish this by doing work on the object, by applying a constant force in the $y$-direction: $$\begin{align} W=F\cdot \Delta y \end{align}$$ It's clear from this equation that applying a force $F=mg$ to the object for a vertical distance of $h$ will result in the precise amount of work required ($mgh$).

If a greater force is applied, you will do "extra" work. This means you will be adding more energy to the object than it needs for the potential energy associated with its height. This "extra" energy will be manifested in the only other available form in this system: kinetic energy.

Conclusion: The total energy of the final state will be the initial state's energy ($0$) plus the work you do. This total energy will be split into kinetic and potential components; the potential energy depends strictly on the vertical position of the object.

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If you give more force, then it will have greater kinetic energy and gets stored as potential energy equally. Potential energy is that an object possess by virtue of its position.

It is independent of the path followed by an object.

In short, it does not gain any energy if you push it with more force or anything. Its merely a measure of potential an object has when placed in a field of force.

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  • $\begingroup$ What do you mean by '''' and gets stored as potential energy equally'''? $\endgroup$ – Paul Jan 12 '15 at 16:09
  • $\begingroup$ more kinetic energy will be converted to as potential energy as the object moves to higher heights, $\endgroup$ – Vinayak Jan 12 '15 at 16:12

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