A math problem!

[yurano]

It's not out of nowhere.


It goes without saying that a = g.

What the hell do you think g is?

g is gravitational acceleration which is a constant.

a is the acceleration you're applying to the object. a can be any number under the sun.

[semaphore]

g is gravitational acceleration which is a constant.

a is the acceleration you're applying to the object. a can be any number under the sun.

Except since we're talking about acceleration due to gravity, a = g. It's honestly pretty damn obvious to, apparently, everyone else in this thread.

[yurano]

Except since we're talking about acceleration due to gravity, a = g. It's honestly pretty damn obvious to, apparently, everyone else in this thread.

And how are you going to prove work done to the object can be converted into PE instead of just work done by the object?

[semaphore]

And how are you going to prove work done to the object instead of just work done by the object?

Your question makes no sense. The solutions presented last page all looks at work done by gravity, hence a = g.

[Deleted]

Do you not just do the triangle formula? When you multiply A and B for example, and end up with C... then when you divide C by B, you are left with A. And the same thing works for C over A. Not sure...

[yurano]

Your question makes no sense. The solutions presented last page all looks at work done by gravity, hence a = g.

So you're going to relegate your answer to only work done by gravity? Instead of an example that incorporates both work done to the system and by the system?

[Monolith of Mazes]

Let's not get into a shouting match over the definition of acceleration. This thread is for the OPs benefit.

There are a lot of ways to define potential energy. You can have gravitational, electromagnetic, spring, chemical, etc. I would assume that, for an introductory physics course, we are dealing with gravity. I asked the OP for his class / teacher / book 's definition of potential energy. Once we have that we can give him a better answer.

[Deleted]

potential U = mgh assuming g is constant. I assume you use W = F*d (non-calc physics?).

F = ma
where acceleration is due to gravity so a = g

distance is the change in height in the gravitational field d = h.
U = (mg)h = F(d) = W

This all also assumes you're talking about gravitational potential. Even if you aren't, you can still show it, but your alphabet soup will look different.

hmm.. i think it is wrong, because d=/=h. Work counts as force * horizontal change of object position. So please remeber that. When you want transport package (f.e 5th floor) using elvator.
Your work will equal as 0.
Because.
W= 10N * 0m = 0J
But when you use the same force while moving 1m. You will get
W= 10N * 1m= 10J

Also I dont know why you used that equation. Problem is simple.
potential energy Ep=mgh
while g=(GM)/(R^2)
M-Earth mass
G- is gravity constant = 6.67

There is no Work when object is falling or elevating. Only sideway move change that. <<< Remember that for life

[semaphore]

hmm.. i think it is wrong, because d=/=h. Work counts as force * horizontal change of object position.

No, it's movement in the direction of the force. I don't know where you got "horizontal change" from.

So you're going to relegate your answer to only work done by gravity?

This seems to be what the OP's question was, so yes.

[Monolith of Mazes]

There is no Work when object is falling or elevating. Only sideway move change that. <<< Remember that for life

That is not correct. To lift something (apply a force for some distance), you must do work against gravity. The whole point of OP's question is to show that a change in potential (i.e. lifting / lowering) is equivalent to work.

This seems to be what the OP's question was, so yes.

Hopefully that's what we're working with. A more general relation might require some calculus :x

[Zedra]

There is no Work when object is falling or elevating. Only sideway move change that. <<< Remember that for life

Sorry but the definition of work done is the product of the force and the displacement in the direction of the force. Since gravity acts downwards, work is done by gravity when an object falls and work is done against gravity when an object elevates. There is no work done by or against gravity when an object moves sideways (horizontally) in a gravitational field.

[Kujako]

Isn't this covered by conservation of energy and the first law of thermodynamics?

[Prokne]

potential U = mgh assuming g is constant. I assume you use W = F*d (non-calc physics?).

F = ma
where acceleration is due to gravity so a = g

distance is the change in height in the gravitational field d = h.

U = (mg)h = F(d) = W

This all also assumes you're talking about gravitational potential. Even if you aren't, you can still show it, but your alphabet soup will look different.

If the problem is to prove that potential energy is a form of work(stored work). Then the solution to the problem uses units.

Basically take the equations and put in the units of the variables.

W=Fxd=m(axd)=kg(m/s^2)m=kgm^2/s^2 which is also equal to a joule(j)

GPE=gph=mgh where g=9.8m/s^2 so mgh=kg(m/s^2)m=kgm^2/s^2=j

You can use any potential energy formula and get the same units as work which means that potential energy can be converted to work.

[Monolith of Mazes]

If the problem is to prove that potential energy is a form of work(stored work). Then the solution to the problem uses units.

Yeah units would work (pun intended? maybe ;D ). You're still using U = mgh F = mg W = Fh, though. You do say that you can use any potential formula and get joules, so yeah unitary analysis would be a better general solution. Depends on what his teacher is defining U as. Hopefully OP is still lurking around and can get some benefit from our tireless pondering.

[Deleted]

Sorry but the definition of work done is the product of the force and the displacement in the direction of the force. Since gravity acts downwards, work is done by gravity when an object falls and work is done against gravity when an object elevates. There is no work done by or against gravity when an object moves sideways (horizontally) in a gravitational field.

Sorry english is not my native languge so you may not understand me properly.

W=F*s*cos(a)

so as you know from math

cos(90)=0

that means that work looks like this

W=F*s*0

so

W=0

OT: But all i can come up with is
( i dont know how to clearly write integral down so i attach img)



So Force depends on the position of points A and B

[Prokne]

Sorry english is not my native languge so you may not understand me properly.

W=F*s*cos(a)

so as you know from math

cos(90)=0

that means that work looks like this

W=F*s*0

a is the angle between the vectors for the force and direction of travel. In the case of an object falling due to gravity a=0 and cos(0)=1.

[Deleted]

Thanks a lot for all of the help I got in this thread, it was all helpful, even though it was initially a little hard to understand most of it. Thanks!

[N-7]

potential U = mgh assuming g is constant. I assume you use W = F*d (non-calc physics?).

F = ma
where acceleration is due to gravity so a = g

distance is the change in height in the gravitational field d = h.

U = (mg)h = F(d) = W

This all also assumes you're talking about gravitational potential. Even if you aren't, you can still show it, but your alphabet soup will look different.

I would solve it like this too.

[Blueobelisk]

Thanks a lot for all of the help I got in this thread, it was all helpful, even though it was initially a little hard to understand most of it. Thanks!

PUH-lease. I just got home and scrolled through the thread and everyone just repeated what I said in a different presentation. Like I said. It was all me. I know it and you know it.

But yeah this question isn't really hard the way you're making it sound, you should be fine.

[Garnier Fructis]

The Work-Energy Theorem states that Work is equal to a change in kinetic energy.

W = delta KE

The Law of Conservation of Energy states that the total energy of a system remains constant.

E = PE + KE = const.

Therefore, a change in potential energy will cause a change in kinetic energy (to maintain conservation). But a change in kinetic energy is equal to work, according to the W-E Theorem. Thus, a change in potential energy is equivalent to work.

That's my version of the solution without the need to invoke any specific types of potential energy or other equations.