# A question about the definition of potential energy with an example

I'm new to Physics and I want to understand everything theoretically and well from scratch.

Let me give you my example.

### Example

Consider now a body (particle), not in intergalactic space, but released from a height h above the surface of the earth. ($$y_{0}=h; V_{0}=0$$) The gravitational force $$F_{G}=-Mg$$ pulls downward on the body. As the body falls toward the surface of the earth, the work done by gravity is equal to the gain in kinetic energy of the body (see Fig. 5.1):

$$W(by gravity)=F_{G} \times (y-y_{0})$$

or, at the surface ($$y=0$$) of the earth,

$$W(by gravity)=(-Mg)(0-h)=Mgh=\frac{1}{2}M{V}^{2}-\frac{1}{2}M{{V}_{0}}^{2}=\frac{1}{2}M{V}^{2} (Eq. 5.7)$$

where $$V$$ is the velocity of the body on reaching the surface of the earth. Equation (5.7) suggests that we may say that at height h the body has potential energy (capacity to do work or to gain kinetic energy) of $$Mgh$$ relative to the earth's surface.

What happens to the potential energy when a particle at rest on the earth's surface is raised to a height $$h$$? To raise the body, we must apply an upward force ($$F_{ag}=-F_{G}$$) to the body. Now $$y_{0}=0$$ and $$y=h$$. We do work,

$$W(by us)=F_{ag} \times (y-y_{0})=(Mg)(h)=Mgh (Eq. 5.8)$$

on the body, thereby giving the body the potential energy $$Mgh$$ that, as we have said earlier, it has at height $$h$$ (see Fig. 5.2a to c). Note that we call the force that we exert $$F_{ag}$$; in other words, we and the external agent are identical. Of course, it is easy to talk about "we" and "us," and the terms are used below; but the important point to remember is that here an external agent is conceptually brought into the problem only for the purpose of evaluating the potential energy.

In the absence of friction forces a specific definition of the potential energy of a body (particle) at a point of interest can now be formulated: Potential energy is the work we do in moving the body without acceleration from an initial loca- tion, arbitrarily assigned to be a zero of potential energy, to the point of interest.

### Question

1) Is Equation 5.8 equals to the change in kinetic energy that is $$\frac{1}{2}M{V}^{2}-\frac{1}{2}M{{V}_{0}}^{2}$$? If it is not, why?

2) In the definition of potential energy, it says that the work we do in moving the body without acceleration. My question is, if we don't accelerate the body how can we raise the body to height of $$h$$ in the example? If we apply equal and opposite force to the gravitational force ($$F_{ag}=-F_{G}$$) to an equilibrium body that has initial velocity of $$V_{0}=0$$, how can we raise the body to the height of $$h$$? If we consider it has a constant initial velocity of $$V_{0}=c$$ we can raise the body to the height of $$h$$, but this time we have a problem with $$W(by us)=F_{ag} \times (y-y_{0})=(Mg)(h-0)=Mgh=\frac{1}{2}M{V}^{2}-\frac{1}{2}M{{V}_{0}}^{2}=\frac{1}{2}M{V}^{2}$$ equation. The body will have constant kinetic energy $$KE=\frac{1}{2}M{V}^{2}$$ and $$U(potential energy)=\Delta(KE)=\frac{1}{2}M{V}^{2}-\frac{1}{2}M{{V}_{0}}^{2}=0$$ so what is the explanation of this? (This question is a little related to my first question.)

Thanks!

if we don't accelerate the body how can we raise the body to height of ℎ in the example?

You can either assume that you accelerate it from rest by a very small amount (so small we can ignore it), or you can assume that it magically begins and ends with some constant speed $$v$$, so no acceleration is required.

but this time we have a problem with $$W(byus) = F_{ag} \times (y - y_0) = (Mg)(h-0) = Mgh = \frac12 MV^2 - \frac12 MV_0^2 = \frac12 MV^2$$equation.

The $$V$$ in the situations are different. In the raising case the work from your hand is going into potential energy, not kinetic energy. So the speed of the object does not change.

In the falling case, all the potential energy is going into accelerating the object (kinetic energy). So the velocity of the object is changing.

You can equate the work done in both cases, but you cannot equate the velocities.

You can calculate the sum of the object's (gravitational) potential energy and kinetic energy (sometimes referred to as the total mechanical energy).

$$E_{tot} = PE + KE$$ $$E_{tot} = mgh + \frac12 mv^2$$

This quantity will remain constant unless there's some work being done on it (by something other than gravity).

In the case where the object is falling, the potential energy is decreasing and the kinetic energy is increasing. At any point in the fall, the sum is constant.

In the case where you are raising the object, the work done by your hand is increasing the total energy. In this case it's going into the potential energy of the object, not the kinetic energy.

• Thanks for you answer. I mean that if the velocity of the object is not changing while it's raising, so there is no change in kinetic energy. So the work we do $W=F_{ag}\times(y-y_{0})=(Mg)(h-0)=Mgh$ and it should be equal to change in kinetic energy. But the velocity of the object is not changing, so the change in kinetic energy is $0$, and $0$ is not equal to $Mgh$. I'm confused :( I think I have got misunderstanding... – ICCQBE Nov 21 '19 at 0:10
• I can read a bunch of brick like books about conservation of energy, energy and work. If you have good and clear resource to learn them well and theoretically, let me know, please. I'm using Berkeley Physics Courses Volume 1 Mechanics book, that you can find online. – ICCQBE Nov 21 '19 at 0:19
• I added a bit to the answer. – BowlOfRed Nov 21 '19 at 0:20
• Oh, wow. That is a good perspective. So while we are raising the object, the system's total energy is changing so the energy is not conserved from the moment we start to raise the object from the zero point. OK, I got it right now. Thank you so much for your time, have a good life! – ICCQBE Nov 21 '19 at 0:30