I What is the zero point for Potential Energy and how to find it from integral limits?

[kirito]

TL;DR Summary

what is the zero point for Potential energy how to find it from integral limits

from what I understand, work is the integral of the force with respect to displacement along the path of displacement $$\Delta w=f*\Delta r$$
and is defined so that the change in kinetic energy is the work,
as for potential energy it is the work done to change the position of the object from point a to b without an acceleration
, it is associated with the placement instead of the path taken, and we are free to choose where the Potential energy is zero,
so if a force is being applied to our object to change its place , we think of a force that is applied in the opposite direction but equal in magnitude,

I am confused about the term zero point where the Potential energy is zero ,
my thought process is the following if a ball is on the ground as soon as I want to calculate the potential I think about its location and where I want it to be and where I want to define the potential energy to be zero ,I think of the force applied to it in the direction I am interested in (gravity ) pulling downwards I choose a force equal in magnitude opposite in direction and name it $F_{ag}=+mg$ $$\int_{0}^{h} F_{ag}$$ will give me the potential energy when it is at the height chosen since I needed to do that much work to move it form point 0 to height h an acceleration , and that would be the potential energy there as for where the potential energy is zero it is on the ground since when I substitute h=0 I get 0 as a result
if I change the limits to $$\int_{0}^{h-y} F_{ag}$$
in my book they wrote that y=0 is the zero point,
I admit assuming its h since the integral would be =0 and work by the force$$ F_{ag}$$ is the potential energy so the work is zero then potential energy is zero
when substituting $$1/2mv^2+mgy=mgh$$ at y=h it is apparent that the kinetic energy of the system is zero It can possibly be zero at the ground if nothing is moving at all or if I am going to throw it from up at height h I assume this is part of why I am confused by also maybe because as I substitute y=0 I get $$1/2mv^2=mgh$$ I start saying this is the potential energy at height h so this is not the point at which potential is zero ,
I know that at y=0 I am at height h and at y=h I am at point h=0
but still lost

 

[Dale]

$$\int_{0}^{h} F_{ag}$$ ...
if I change the limits to $$\int_{0}^{h-y} F_{ag}$$

Your change of limits is incorrect. Also you are forgetting the differential, and forgetting to write clearly what variable $F$ is a function of. Basically, you are being very sloppy and your confusion is a result of that sloppiness. Set up your variables clearly and do your substitutions clearly.

So, let's suppose that you have a force $F (x)$, if you want to integrate that force along the path from $x=0$ to $x=h$ then you write $$\int_0^h F(x) \ dx$$

Now, if we want to change the limits of integration then we can do something like the substitution $y=x+H$. This gives $x=y-H$ and $dx=dy$. Then the integral is $$\int_H^{h+H} F(y-H) \ dy$$

This will work out correctly. Just be careful with your math.

 

[kirito]

Your change of limits is incorrect. Also you are forgetting the differential, and forgetting to write clearly what variable $F$ is a function of. Basically, you are being very sloppy and your confusion is a result of that sloppiness. Set up your variables clearly and do your substitutions clearly.

So, let's suppose that you have a force $F (x)$, if you want to integrate that force along the path from $x=0$ to $x=h$ then you write $$\int_0^h F(x) \ dx$$

Now, if we want to change the limits of integration then we can do something like the substitution $y=x+H$. This gives $x=y-H$ and $dx=dy$. Then the integral is $$\int_H^{h+H} F(y-H) \ dy$$

This will work out correctly. Just be careful with your math.

sorry I did forget that the integrations I wrote where for the following instructions and just wrote them as it is without any details and your comment shed light on that , you have my thanks
in my book they started by providing us with the formula $$1/2mv^2-1/2mv_0^2=1/2mv^2 =mgh $$
$$\int_0^{h}mg \ dr=\Delta K_E$$
then they said if v is not the velocity at height h instead at height h-y then
1/2mv^2=mg(h-y)
$$\int_{0}^{h-y}mg \ dr=\Delta K_E$$

and then they stated y=0 gives us mgh as when so y is the zero point
so I still fail to understand this even with the right limits

 

[Dale]

This makes no sense:

$$\int_0^{h}=mg\delta r=\Delta K_E$$

$$\int_0^{h}= \ ...$$ is not a well formed expression. If that really is the expression in your book then you need to get a new book.

You need to not be sloppy. Also, define your terms here. I assume $m$ is mass and $g$ is the gravitational acceleration. What are $h$ and $\delta r$ and later on what is $y$?

 

[kirito]

This makes no sense:
$$\int_0^{h}= \ ...$$ is not a well formed expression. If that really is the expression in your book then you need to get a new book.

You need to not be sloppy. Also, define your terms here. I assume $m$ is mass and $g$ is the gravitational acceleration. What are $h$ and $\delta r$ and later on what is $y$

sorry still new to latex so all of the symbols formatted differently from what I intended took me a while to see how to fix them integration in respect to really change in r such that r is a symbol for the displacement , m is mass , g is gravitational acceleration h is the max height , and if I understand this correctly y is a value smaller or equal to h so that y is considered the height with the ball is at and sum of them is a constant, I seem to have reached an understanding while writing this

 

[Dale]

then they said if v is not the velocity at height h instead at height h-y then
1/2mv^2=mg(h-y)

That is correct, but then this is not a change of variable, this is a completely different integral so it does not make sense to say that they are both equal to the same $\Delta KE$

 

[kirito]

That is correct, but then this is not a change of variable, this is a completely different integral so it does not make sense to say that they are both equal to the same $\Delta KE$

oh about that it is stated above that$$ \Delta ke$$ in the second equation is for velocity at height y so that it has a different velocity excuse my sloppiness I will learn try to highlight such details in a better manner in the future