# What is the derivation of potential energy of an object?

I want to know why we use the gravitional force to compute the potential energy.

Assume we have a ball, and we lift it up by a distance $$h$$. Then, as I understand, the force $$F_l$$, must have been more than the gravity force, $$Fg$$. Other than this, the ball would not have moved;

$$PE = \underbrace{m.g}_{Fg < F_l}.h$$

Assuming this, why do we use the gravitional force to compute the potential energy, while the force that lifts the object must be more than that force in order to move?

I think the potential energy would be something like $$PE = \underbrace{F_l}_{F_l > m.g}.h$$

• I want to know why we use the gravitional force to compute the potential energy. This is backwards. There isn't a general "potential energy" and we use the gravitational force to find it. Rather, the force of gravity has a potential energy associated with it, and we can use the idea of gravitational potential energy instead of explicitly considering the work done by gravity. This is independent of other forces acting on the object. – Aaron Stevens Jan 29 at 17:53

While it is true we initially need to apply a force slightly greater than gravity to begin to lift the mass, as soon as we start it moving we can reduce our force to to exactly equal mgh. Then, since the mass now has some kinetic energy, we need to reduce the force to slightly less than gravity so the mass decelerates giving us a negative change in kinetic energy to cancel the initial increase in kinetic energy when it stops at the height h. Then the only energy the mass has is potential energy, $$PE=mgh$$.

Now with regard to your equation:

$$PE=f_{l}.h$$

Where $$f_{l}>mg$$, if that force remains applied to the mass up to when it reaches height $$h$$, then you will have kinetic energy as well as potential energy when the mass reaches $$h$$. The kinetic energy portion will be, by the work-energy principle,

$$KE=\frac{mv^2}{2}=(f_{l}-mg)h$$

So your equation should not be all PE but PE+ KE and should be

$$PE+KE= mgh+(f_{l}-mg)h$$

Hope this helps.

• "While it is true we initially need to apply a force slightly greater than gravity to begin to lift the mass..." - Not in the case where the object already has an initial velocity. It doesn't seem to be assumed in the question that the object is at rest. – probably_someone Jan 29 at 17:46
• @probably_someone I was assuming the mass starts out at rest. I'll make that clarification. Thanks – Bob D Jan 29 at 17:48

You have stated the potential energy as the product of the external force and the distance moved by the object.It is not correct to say that.Firstly Potential energy is the negative of the work done by the conservative force , which is the gravitational force in this case , from your question I believe you have not been introduced to variable forces and conservative forces , so i am not going to dwell on that , for now the force is always mg above the surface of earth , pulling the body downwards
You can think of potential energy only with respect to a reference point , the reference point is the surface of earth here , As you should be able to make out from the term itself , it's the POTENTIAL energy , something which is yet to be expressed , so think along this scenario-If you take the object at a certain height above the surface and drop it how much kinetic energy will it gain by the time it strikes the ground,you can do that easily , let h be the height above the surface where you took the object $$v^2=2gh \Rightarrow KE=\frac{1}{2}m(2gh)=mgh$$which gives you the value of potential energy of the ball at that height.
Now you have also asked why have we lifted the object with the force mg.The minimum force required to move an object in the opposite direction on which a force F is acting is to apply a force F in the exact opposite direction,at this instant if the body starts to move in the desired direction it will continue to do so indefinitely since the net force on it is 0, If some force greater than mg was taken then there would be a net force on the object in the upward direction and by the time you would have pulled it to h height it would have gained certain velocity in the upward direction which will cause it to move to some greater height H after you turned off your external force at h , from that greater height when the object falls the Kinetic energy upon striking ground will be more as compared to Kinetic energy when it fell from h therefore it is safe to say that potential energy in the former case was more , but you wanted to calculate the potential energy at h height and not at H , therefore force used is exactly equal to gravitational force , to avoid building up Kinetic energy in the pulling up.

Hope this helps,quite an explanation!

There is an incorrect assumption here:

Assume we have a ball, and we lift it up by a distance $$h$$. Then, as I understand, the force $$F_l$$, must have been more than the gravity force, $$F_g$$. Other than this, the ball would not have moved.

Forces are the cause of acceleration, not motion. Remember Newton's First Law: in an inertial frame, an object not acted on by forces will continue to move at the same velocity. So an object that has a nonzero initial velocity, where $$F_l=F_g$$, will continue to move at that velocity, and, if that velocity is directed upward, after a certain amount of time will reach a height $$h$$.

At this point, you can say that the lifting force did an amount of work equal to $$F_l h$$ on the object; it increased its potential energy by $$F_l h$$, and its kinetic energy did not change (it's still moving at the same speed). You can also say that the gravitational force did work $$-F_g h$$ on the object, since the force was oppositely directed to the motion. Since $$F_l=F_g$$, the net work on the object is zero, which is consistent with the work-energy theorem:

The total work done on an object is equal to the change in kinetic energy of that object.

Let's assume now that $$F_l>F_g$$. Now the object accelerates upwards; its kinetic energy is higher at height $$h$$ than it was originally. How much kinetic energy did it gain? Well, the work-energy theorem tells us: the amount of kinetic energy it gains is precisely equal to the net work on the object, which is $$(F_l-F_g)h$$. Indeed, this makes all of the energy totals add up correctly; the lifting force inputs a total energy of $$F_lh$$ into the gravitational system. Of this energy, an amount $$F_gh$$ went to the potential energy of the object, and an amount $$(F_l-F_g)h$$ went to increasing the kinetic energy, so everything is accounted for.

To answer the broader question of why we use the gravitational force for the gravitational potential energy, we need to define potential energy for an arbitrary conservative force. In general, the potential energy is the energy contained in the configuration of a system. What "configuration" means can differ for different forces; in the particular case of the gravitational force, this is the energy associated with the relative positions of everything in the system. Since, for our purposes, the Earth is fixed, the gravitational potential energy is the energy associated with the height of the object.

The other key piece of understanding is this: the work done by a given conservative force is equal to the negative change in potential energy of the object. (This comes from the more fundamental statement that force is the gradient of potential energy). Since gravity is a constant force $$mg$$ in the approximation you're using, the work done by gravity is simply the force multiplied by the distance the object moves in the direction of the force, which is the definition of height. From this you get that the work done by gravity is $$-mgh$$, so the associated change in gravitational potential energy is $$mgh$$. If we choose to set the gravitational potential energy of the object at $$h=0$$ to be $$PE=0$$, then we have our formula $$PE=mgh$$.