## Physics Energy Question

For the discussion of the sciences. Physics problems, chemistry equations, biology weirdness, it all goes here.

Moderators: gmalivuk, Moderators General, Prelates

Posts: 14
Joined: Mon May 26, 2008 4:37 am UTC

### Physics Energy Question

hey, sorry to post with only an intro under my belt but hey,
also sorry if this is a really stupid question.

OK consider the situation of a mass (m) on a spring. the spring compresses by the distance (x). it is now at rest, not oscillating.

therefore
Ep Lost = Ep Gained
mgx = 1/2kx2
mgx = 1/2Fx2
x
mgx = 1/2Fx
mgx = 1/2mgx

EDIT: Spelling
EDIT2: changed name on request
Last edited by Phaden on Tue May 27, 2008 9:28 am UTC, edited 1 time in total.

Charlie!
Posts: 2035
Joined: Sat Jan 12, 2008 8:20 pm UTC

### Re: Physics

You need to recheck your math. You definitely have the right idea, though.

hint:
mgx = 1/2kx2 --> mg = 1/2kx
Some people tell me I laugh too much. To them I say, "ha ha ha!"

Gammashield
Posts: 68
Joined: Fri Jun 29, 2007 7:41 pm UTC

### Re: Physics

The problem with your logic is subtle, actually. That little note: "the spring compresses a distance x. it is now at rest, not oscillating"

You arrive at a contradiction at the end (1/2 mg = mg) and so, yeah, something went wrong, but there aren't any errors in your math (er, though you might want to reformat that third line. It took a moment to realize what you were trying to say). The mistake is the first line, your initial assumptions.

The energy stored in the spring at the 'balance point', where Fg =Fs (force of gravity = force applied by the spring) is *not* the energy gained by lowering the mass m that distance. Let me try to show why. Imagine the physical experiment. You have your spring in its rest position, at x=0. You stick a mass on it. Ignore friction, air resistance, etc etc; we're assuming no energy loss. What happens?

The spring compresses down some amount, reaches the balance point, but is still moving. It has kinetic energy at the balance point. It then continues compressing the spring, and oscillates around the balance point continuously. At the balance point, in fact, half the energy gained by lowering it to that point graviationally is potential energy in the spring. half is kinetic energy.

We *can* stop the spring from oscillating, but in order to do this we have to apply some force to it; friction or otherwise, and by doing so we pull energy out of the system. And after that, we can't expect the initial gravitational energy to equal the potential energy in the spring.

So, yeah, mathematically, the error is saying "Ep lost = Ep gained." In stopping the oscillation, we had to pull energy out of the system, so we can't make that statement any more.

Hope that helps.

Posts: 14
Joined: Mon May 26, 2008 4:37 am UTC

### Re: Physics

meh, i knew there had to be a problem with it somehow.
It had me stumped though and my physics teacher did not know what was wrong either.

(im only in year 12 physics though)

just a question wouldn't the oscilations decrease in size over time and this would tend to the equation of mine?

as a side note, that 3rd line is misleading but the x should be under the rest of the fraction. (dont know how to move it as when i put spaces in it moves it back anyway.

Charlie! wrote:You need to recheck your math. You definitely have the right idea, though.

hint:
mgx = 1/2kx2 --> mg = 1/2kx

where? though you might also be confuzzeled with the 3rd line.
thats a cool word
confuzzeled.....

Gammashield
Posts: 68
Joined: Fri Jun 29, 2007 7:41 pm UTC

### Re: Physics

Phaden wrote:meh, i knew there had to be a problem with it somehow.
It had me stumped though and my physics teacher did not know what was wrong either.

(im only in year 12 physics though)

just a question wouldn't the oscilations decrease in size over time and this would tend to the equation of mine?

Oh, certainly no shame in being stumped for a moment on that one; like I said, it's a fairly subtle point, since the math looks pretty darned convincing, and free-body diagraming just seems to confirm the whole thing. I could see a teacher who's not checking his assumptions too carefully getting tripped up.

The oscilations *do* indeed decrease in size over time, but for a reason. Friction on the air. Friction that causes heat, causes energy loss, etc. These are all measurable forces, that cause measurable amounts of energy to leave the system. So the proper equation would be " (Ep lost)= (Ep gained) + (E lost to friction)" If you did the experiment in a perfect vacuum, then actually, no, the oscilations *wouldn't* decrease over time. One of those weird, non-obvious effects of a vacuum, sort of like the feather-and-bowlingball falling trick.

Whenever one is dealing with an energy-conservation question, it always pays to be *very* careful about how energy might leave the system. It can often be in small, unexpected ways, even in a purely theoretical, perfect-spring setup like this.

scowdich
The Hedgehog
Posts: 771
Joined: Tue May 22, 2007 4:55 am UTC
Location: University of Illinois (Urbana-Champaign)
Contact:

### Re: Physics

These oscillations would tend to decrease over time, but this is due to air friction/heat losses in the spring from compression/various other squidge factors. When you're doing conservation-of-energy calculations like these, We Can't Have That Sort Of Thing Around Here - everything's perfectly spherical, massless, frictionless, and moving through a vacuum unless explicitly stated otherwise.

Dammit, ninja'd.

Posts: 14
Joined: Mon May 26, 2008 4:37 am UTC

### Re: Physics

i hate those ninjas

Posts: 14
Joined: Mon May 26, 2008 4:37 am UTC

### Re: Physics

sorry to rebring this up but i was just thinking on this today and still dont fully get it.
at the bottom of the oscillations the kinetic energy is 0 for a moment in time and this formula becomes true.....
doesn't it?

to illustrate what i mean, if you are walking forward and then change and walk backwards for a brief moment of time your velocity is 0 and your kinetic energy is 0

evilbeanfiend
Posts: 2650
Joined: Tue Mar 13, 2007 7:05 am UTC
Location: the old world

### Re: Physics

yes at the bottom of the oscillation the velocity and KE is zero but the PE is the spring is at a maximum, the mass is instantaneously not moving but is still accelerating (:. still has a force on it :. not in equilibrium). likewise, at the equilibrium point the mass will be moving but not accelerating (unless damping has reduced the movement here to 0 in which case your system has stopped in equilibrium)

your mistake seems to be equating 0 velocity with being in equilibrium but this is not the case. newtons laws explicitly state that an object at constant velocity is also in equilibrium the key here is there must be no net force/acceleration on the object for the object to be in equilibrium.

edit: also is there any chance you can make your subject a little more descriptive? "physics" covers pretty much 80% of all threads in this subfora.
in ur beanz makin u eveel

Posts: 14
Joined: Mon May 26, 2008 4:37 am UTC

### Re: Physics

evilbeanfiend wrote:yes at the bottom of the oscillation the velocity and KE is zero but the PE is the spring is at a maximum, the mass is instantaneously not moving but is still accelerating (:. still has a force on it :. not in equilibrium). likewise, at the equilibrium point the mass will be moving but not accelerating (unless damping has reduced the movement here to 0 in which case your system has stopped in equilibrium)

your mistake seems to be equating 0 velocity with being in equilibrium but this is not the case. newtons laws explicitly state that an object at constant velocity is also in equilibrium the key here is there must be no net force/acceleration on the object for the object to be in equilibrium.

edit: also is there any chance you can make your subject a little more descriptive? "physics" covers pretty much 80% of all threads in this subfora.

Not sure i understand what this has to do with the original equation as it is talking about the transfer of energy. i didn't say it had no resultant force i said it had no KE. where did the rest of the energy go????

on the name, sure i could give it a go but is it really necessary? once someone has read it they will know what it is dont they?

EDIT: Changed name and did sum spelling checks

evilbeanfiend
Posts: 2650
Joined: Tue Mar 13, 2007 7:05 am UTC
Location: the old world

### Re: Physics

Phaden wrote:Not sure i understand what this has to do with the original equation as it is talking about the transfer of energy. i didn't say it had no resultant force i said it had no KE. where did the rest of the energy go????

its stored in the spring, the key you are missing is your system at the lowest point of the oscillation has no KE but is not in mechanical equilibrium.

edit: unless you are considering damping but we have already discussed that, there is no KE then as we have been losing energy to friction etc. and we are in mechanical equilibrium but we aren't at the lowest point of the oscillation.
Last edited by evilbeanfiend on Tue May 27, 2008 9:50 am UTC, edited 1 time in total.
in ur beanz makin u eveel

Posts: 14
Joined: Mon May 26, 2008 4:37 am UTC

### Re: Physics Energy Question

which is???
remember from up top im only y12 so a few things may need to be explained a little
y does equilibrium effect energy?

evilbeanfiend
Posts: 2650
Joined: Tue Mar 13, 2007 7:05 am UTC
Location: the old world

### Re: Physics Energy Question

hmm have you drawn pictures of your system at the start and end? that can help.

otherwise see if you agree with these:

a1) initially the mass is dropped above the spring from x=0, it has forces on it
a2) there is no damping
a3) the mass oscillates forever. while it has 0 velocity at points it is not in equilibrium at these points :. not at rest (assuming 'at rest' means 0 velocity and 0 forces). at the equilibrium point the mass has maximum KE.

or

b1) initially the mass is dropped above the spring from x=0, it has forces on it
b2) there is damping
b3) the mass oscillates for a finite time but eventually all KE at the equilibrium point is lost. the mass is then at rest. the missing energy has either been taken out of your system completely (your system only considers mechanics and is :. not closed) or basically ends up being heat, your equipment is now warmer.

c) there is no way to get the system at rest without damping unless the system starts at rest (in which case x=0 and it isn't very interesting)

edit: in terms of energy, equilibrium will mean total PE = 0, it says nothing about KE at all.
Last edited by evilbeanfiend on Tue May 27, 2008 10:42 am UTC, edited 1 time in total.
in ur beanz makin u eveel

tdc
Posts: 14
Joined: Fri May 02, 2008 2:22 pm UTC

### Re: Physics Energy Question

The mechanical equilibrium is the point where, if you place the mass in it, it will not move away - all forces on it cancel out.
So the easiest way would be to equate the two forces. (mg = kx)

If you really want to use Energies, you can also do that. Let U be the total potential energy of the system (U=-mgx + 1/2kx^2)*.
Equilibrium points will satisfy U' = 0 (mg = kx), i.e. are extrema of the potential. **
You can then use U'' (in this case U'' = k) to find out whether you have a local maximum or a local minimum (here: minimum). A minimum would be called a stable equilibrium, which means that if you slightly displace the mass, it will start to oscillate around the equilibrium point. A maximum conversely would be instable - if you slightly displace the mass it will move away from the equilibrium point.

* The mgx term is negative because we've defined x to be in the downwards direction by saying that the spring is compressed by x.
** A good way of getting a feeling for this is to draw a graph of the potential over x and then pretend it's a slope and set a ball on it. Will it move?

Mettra
Posts: 96
Joined: Tue Apr 29, 2008 1:37 am UTC

### Re: Physics Energy Question

remember from up top im only y12 so a few things may need to be explained a little
y does equilibrium effect energy?

Think about it like this. Draw two free body diagrams - one at the initial conditions and the other when the mass has been displaced.

Since you are using mg, I assume the spring is vertical and the mass is hanging downwards from it. If not, this method will still work either way.

At the initial starting point, you draw the mass' free body diagram with the force of the spring Fs pointing up and the force of gravity Fg pointing down. Don't concern yourself with numeric or algebraic quantities yet. Consider the magnitudes of the two forces acting on the mass. If Fg is bigger than Fs, the mass will want to be accelerated downward (remember, where there are forces, there are accelerations). On the other hand, if Fs is bigger than Fg, the mass will want to accelerate upward. The only way that the mass won't accelerate is if both forces are exactly equal. So initially, the two forces must be equal (since you said it was at rest).

After the displacement, you are right that instantaneously there is no motion (if you pause the 'video', it will just sit there). However, draw the FBD and you will see that one of the forces is now larger than the other force. This will cause an acceleration.

This acceleration causes the kinetic energy (and is caused by the force). All of this stuff is very closely related, though you won't use it for this specific problem.

Your statement "i didn't say it had no resultant force i said it had no KE. where did the rest of the energy go????" tells me that you have forgotten a special quantity called potential energy.

The work-energy theorem gives you Wnet = Kfinal - Kinitial

Since this is an equation, you can transform it at your will, and we have defined in physics a very useful quantity by doing this. We'll say that some quantity (Ufinal - Uinitial) is equal to minus the net work.

-(Uf - Ui) = Wnet = Kf - Ki

-(Uf - Ui) = Kf - Ki

0 = (Uf - Ui) + Kf - Ki

Or more generally

Wncf = (Uf - Ui) + (Kf - Ki)

where Wncf is the work done by Non-Conservative Forces (friction)

-----
You can see that deltaU and deltaK (can't get JSmath to load for some reason) always have to equal 0 if there aren't any NCF's. That means if you lose some KE, you will gain some U. Furthermore, if you gain some KE, you will lose some U. This quantity U is called the potential energy, and we generally imagine it as being 'stored' energy waiting to convert into KE. For example, think of the spring that the mass is hanging on. If that mass suddenly disappeared, what would the spring do? It would pull back in. That's where the potential energy is stored.

With the above equation you should be able to do and understand just about any problem with work.
zenten wrote:Maybe I can find a colouring book to explain it to you or something.

Gammashield
Posts: 68
Joined: Fri Jun 29, 2007 7:41 pm UTC

### Re: Physics Energy Question

remember from up top im only y12 so a few things may need to be explained a little
y does equilibrium effect energy?

Hrm.... I think the simplest way to say it is that, yeah, equilibrium doesn't affect the total energy in the system (just how things are moving from potential to kinetic and back, which I assume you understand just fine). If we've got that oscillating body with no friction, all the energy is always in the system, no matter what form it's in, no matter whether or not the system in equilibrium. It is important not to lose that point in the discussion of how to deal with and watch the energy. Conservation of energy does *not* depend on equilibrium, be careful not to misread anyone's posts as saying that. But being in non-equilibrium states does mean we have to watch which substitutions and equalities we make.

Phaden wrote:sorry to rebring this up but i was just thinking on this today and still dont fully get it.
at the bottom of the oscillations the kinetic energy is 0 for a moment in time and this formula becomes true.....
doesn't it?

to illustrate what i mean, if you are walking forward and then change and walk backwards for a brief moment of time your velocity is 0 and your kinetic energy is 0

I'll take my shot at answering this, since I'm not sure if it's clicking for you yet. I may not help any more than the other explanations you've been getting, but heck, I like hearing myself talk.

At this point, the bottom of the oscillation, your line of equations has a *different* problem. The first line is actually true now. If there's no kinetic energy, and no loss to friction, yeah, everything that was gravitational is now potential energy in the spring, so mgx = 1/2kx2

However,

Look at what you do with F in that proof. First in step two you say F = kx, and run the substitution. Later in step 4 you say F = mg , and run *that* substitution. As everyone's mentioned, at the bottom of the oscillation the forces *don't* balance, since yeah, no equilibrium, there is still a net force. The force of the spring, kx, does not equal mg. That's now an illegal step.

As a matter of fact, you can work out what the forces are at this point and go through your proof as it works at that bottom point of the oscilation and see what happens. Let's say the bottom of the oscillation is some distance x down (sidenote: remember to be careful about your variables. This is a different x than the 'distance x to the balance point' I was talking about in previous posts)

So, the spring compresses a distance and comes to a stop at the bottom of the oscillation (soon to spring back up, but heck, we don't need to worry about what it will do later, we're just worried about that instant when it stops, now)

let's get a few statements down first:

Egained = (1/2)*kx2
Fspring = kx

Elost = mgx
Fgravity = -mg (minus sign, since gravity's force is downwards)

How are Fs and Fg related, though? They're not equal, the situation isn't in equilibrium, so the forces don't match. Instead, there's some net difference. If you think about the symmetry of the situation (heh, there's a physics term you'll hear more and more as time passes. symmetry.) We can realize that the net force at the top of the spring is equal and opposite to the net force at the bottom of the spring. The net force at the top of the spring comes from gravity only (no spring compression at the top of the oscillation in this problem) *So*, we know that

Fnet = mg (positive, since at the bottom it's equal and *opposite* to what it is at the top)

Fs + Fg = Fnet
kx - mg = mg
(1/2)kx = mg

now that we've worked out that relationship at the bottom of the oscillation, we can run through your proof. We can now define F = (1/2)*kx = mg at the bottom of the oscillation.

Elost = Egained
mgx = (1/2)*kx2
mgx = (1/2) * (2*F/x) * x2
mgx = Fx
mgx = (mg) *x
mgx = mgx

Constants equal themselves, physics works, and all is right with the world.

Posts: 14
Joined: Mon May 26, 2008 4:37 am UTC

### Re: Physics Energy Question

Gammashield wrote:Look at what you do with F in that proof. First in step two you say F = kx, and run the substitution. Later in step 4 you say F = mg , and run *that* substitution. As everyone's mentioned, at the bottom of the oscillation the forces *don't* balance, since yeah, no equilibrium, there is still a net force. The force of the spring, kx, does not equal mg. That's now an illegal step.

thanks, thats what i forgot and was very easy to understand 2, tyvm

Gammashield wrote:Constants equal themselves, physics works, and all is right with the world.

it would be fun to find a situation where physics fails though......