A math problem!

[Deleted]

So, being the uneducated schmuck that I am, I've gotten stuck on this (part of an) assignment that I have, due for tomorrow.
Basically, it reads (translated):

"Prove that a change of potential energy is a work. Basically you should show that a change in potential energy equals force times distance."

Yeah, it's weirdly worded, but it's not much better in Swedish (I believe the teacher is dyslexic).

Anyone who can give me a hand?

[Ryme]

So, it's been about 6 years since I did physics, but this sounds to me like you're just trying to prove an equivalence relation, mess around with one side of the formula until it is equivalent to the other.

[Deleted]

Oh yeah, physics is what the subject is called. I mixed them up as I'm doing math at the same time. Oops.

You mind explaining what you mean by "mess around with one side of the formula"?

[Blueobelisk]

this is physics.

Potential energy = mgh for gravitational potentional. M being mass, g being the gravity acceleration constant, which is a force. h is depending on the height, which is the distance.

and as you said work = force*distance

then again you could always integrate a work chart but that seems hard. I don't really know how else to help you.

---------- Post added 2013-01-28 at 08:51 AM ----------

I have to go to class now so i'll check this thread later, if you need more just let me know.*COUGH* Although some dorks will show up and basically repeat what I say and make it seem like they did something *COUGH COUGH*

[Deleted]

Yeah, I realise my mistake. I should've named it "A physics problem!" (if a mod wants to change it, feel free). Oh well.

My problem is that I'm not sure what the teacher actually wants out of this question, since he explained the formula. I guess I need to go into details to explain and prove why it works like that, and I'm honestly not sure exactly *why* force times distance equals work. >_<

[Monolith of Mazes]

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.

[Deleted]

I assume you use W = F*d (non-calc physics?).

Yeah, I do.

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'm pretty sure I'm not. The word in Swedish is lägesenergi which directly translated means location energy.

I'm ashamed to admit that your formula makes little sense to me as I haven't done physics in many years and I'm now taking distance courses to prepare for university this autumn. I'm as rusty as can be, and the last physics I did was in 9th grade (5 years ago).

[Monolith of Mazes]

Okay. Force is mass * acceleration, F= ma.

I don't know what kind of definition you're given for potential energy, U.

Work, W, is a force being applied for some distance, W = Fd.

Can you provide the definition of potential that you're given? Book, class notes?


Edit: off to class, I'll look into this thread again when im done

[Reg]

E = mc^2

Did I physics right?!

So glad I went a different route with my college career lol

[semaphore]

U = m.g.h
W = F.d
F = m.a

W = -m.g.d
W = -m.g.Δh
W = -ΔU

I'm honestly not sure exactly *why* force times distance equals work. >_<

F = m.a
F.d = m.a.d
d = v₀.t + 0.5.a.t² ∴ F.d = m.a.v₀.t + 0.5.m.a².t²
a = (v - v₀)/t ∴ F.d = m.v₀(v-v₀) + 0.5.m(v²-2v.v₀+v₀²)
F.d = 0.5.m.v² - 0.5.m.v₀² = W

I think. But I wouldn't trust my math, or physics, or brain.

[yurano]

Are you sure you're trying to prove that work is related to a change in potential energy? Work done on a system adds energy into the system, increasing 'total energy'. Total energy = kinetic energy + potential energy. It could be that the change in total energy is non zero, yet the change in potential energy is zero.

If the question was instead how to prove that work can be related to a change in kinetic energy, then the proof is shown in the following video starting from 3:50.

https://www.khanacademy.org/science/...ork-and-energy

[semaphore]

Are you sure you're trying to prove that work is related to a change in potential energy?

It reads to me like they're asking the OP to show that a change in potential energy = work done by gravity. But I dunno

[yurano]

It reads to me like they're asking the OP to show that a change in potential energy = work done by gravity. But I dunno

If you're talking about converting work into gravitational potential energy, you'd use equation 6.7 from the following PDF: http://iweb.tntech.edu/murdock/books/v1chap6.pdf

Lets assume for the moment that we have an object of mass m at y = y1 (height). We are going to apply a force F that exactly counteracts gravity over a distance d, raising the object by d.

According to our work function, W = integral(F(y), y1, y1 + d). Since F(y) = m * g is 'constant' with respect to y, our equation becomes W = m * g * d. We see that we've raised the object a height d which would increase gravitational energy by m * g * d which is exactly how much energy was applied as work.

This explanation makes one of two assumptions (probably irrelevant for the scope of the question):

1) the object was already at motion (however infinitesimally) in the direction we applied our 'work'. Since our force F exactly counteracted the force of gravity, the object did not accelerate/decelerate for the duration of our 'work'.

2) the object was at rest originally and accelerated at the beginning (consuming additional energy) and decelerated at the end ('additional' consumed energy is recovered).

[jonny888]

Slightly modifying what Semaphore described

U = m.g.h
W = F.d
F =m.a

Substituting for F
W = m.a.d

In this case
a = g

So
W = m.g.d

For your purposes
d = h

Therefore

W = m.g.h
and
U = m.g.h

resulting in U = W

If U1 is starting potential and U2 is finishing potential energy
U1 - U2 = W1 - W2

Thereby showing that a change in potential energy can be represented by a change in the work done
I think this is what you're after?

[ripponesan]

Well if there was no gravitiy there would be no potential energy, so it is safe to assume gravitiy can be included in this lesson.

One could argue that a stretched rubberband holds potential energy, and you are kinda right but physicists have a own word for that kind of energy, so yeah potential energy directly relates to gravity.

---------- Post added 2013-01-28 at 04:31 PM ----------

Slightly modifying what Semaphore described

U = m.g.h
W = F.d
F =m.a

Substituting for F
W = m.a.d

In this case
a = g

So
W = m.g.d

For your purposes
d = h

Therefore

W = m.g.h
and
U = m.g.h

resulting in U = W

I think this is what you're after?

Quite an evolved approach and still understandable for starters i recall. great.
Only thing to keep in mind is that this describes ideal circumstances where you dont have any losses through friction.

Tell that to your teacher, he probably will love it.

[semaphore]

If you're talking about converting work into gravitational potential energy, you'd use equation 6.7 from the following PDF: http://iweb.tntech.edu/murdock/books/v1chap6.pdf

I'm saying it sounds like they're talking about 6.18. This assuming a new force and stuff kinda just overcomplicates things.

[yurano]

I'm saying it sounds like they're talking about 6.18. This assuming a new force and stuff kinda just overcomplicates things.

6.18 doesn't 'prove' that you can convert work into PE and visa versa. 6.18 leads from the previous equations such as 6.7 and the answer to the OP's question.

Here's the most complete way to answer the question if we're talking about GPE:

A object of mass m at height h0 has a gravitational potential energy of U0 = m * g * h0 (1)

We apply a force F equivalent to gravitational force on the object over a distance d. (F = m * g) Work done to the object W = F * d; W = m * g * d (2)

The object is now at height h0 + d and has a gravitational potential energy of h1 = m * g * (h0 + d) (3)

From (1) and (3), the change in potential energy ΔU = m * g * d which is equivalent to the work done to the object (2).

The problem with semaphore's and jonny888's solutions is the assertion that a = g. The equation a = g is not 'given' nor a 'law' so you can't assert it without reasoning. Both of solutions are just manipulations of variables instead of 'proving' with model of how work is converted in GPE.

[semaphore]

6.18 doesn't 'prove' that you can convert work into PE and visa versa.

Who said convert what? The point is that the change in potential energy is equal to work done by gravity.

---------- Post added 2013-01-28 at 03:57 PM ----------

The problem with semaphore's and jonny888's solutions are the assertion that a = g. The equation a = g is not 'given' nor a 'law' so you can't assert it without reasoning.

You're just being ridiculous at this point.

[yurano]

Who said convert what? The point is that the change in potential energy is equal to work done by gravity.

Using 6.18 is jumping the gun. The whole point of the problem is to prove 6.18 (via GPE).

You're just being ridiculous at this point.

You can't pull a = g out of nowhere. Why not a = 2g? Why not a = 3g? Why a = g? "Because it fits the equation" isn't a valid reason. In fact, you gave no reasoning given at all for why a = g.

[semaphore]

You can't pull a = g out of nowhere. Why not a = 2g? Why not a = 3g? Why a = g?

It's not out of nowhere.

In fact, you gave no reasoning given at all for why a = g.

It goes without saying that a = g.

What the hell do you think g is?