Energy not conserved?

[andafoo]

A vertically oriented spring is stretched by 0.50 m when a mass of 1 kg is suspended from it. What is the work done on the spring?

A. 5.0 J
B. 2.5 J
C. 0.5 J
D. 20.0 J

Correct Answer: B
Explanation


At equilibrium the force of the weight (W = mg) will be equal to the spring force (Fs = -kx), thus:
kx = mg, k = mg/x = (1 kg x 10 m s-2)/(0.5 m) = 20 kg s-2

Work done (spring) = 1/2 kx2 = 1/2 (20)(0.5)2 = 1/2 (20)(1/4) = (20)/(8) = 5/2 = 2.5 J.



My question is, why is the energy stored in the spring not equal to the loss in potential energy of the block?

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[mosquitoman]

some energy is dissipated in the spring as heat

pe = mgh gives you 10*1*.5 = 5J

take the initial height [before release to be .5m off the ground]

so 2.5 J is lost as heat

I believe this is correct

 

[Ninjaface]

My question is, why is the energy stored in the spring not equal to the loss in potential energy of the block?

The potential energy of the spring due to the block is equal to that lost by removal of the block though - it is all caused by the block being there, and is all gone when the block leaves, no? Maybe I am not understanding your problem sorry. The above response is definitely incorrect though (no offense mosquitoman), for one h above the ground is not given, and for another you can't just make up a value for heat because you want it...

 

[andafoo]

I was wondering if this had to do with conservative/nonconservative forces... which I don't quite the best grasp of.

To clarify my question, I'm assuming conservation of energy. Assuming the system to be the spring and block together, the total system should not increase or decrease in total energy?

So when the block is lowered, it goes from a higher absolute height to 0.5m below that - resulting in a decrease in gravitational potential energy. Where do those 5J of potential energy go? Well I currently can only account for the energy converted into elastic potential energy in the spring... where's the rest go?

 

[mosquitoman]

The potential energy of the spring due to the block is equal to that lost by removal of the block though - it is all caused by the block being there, and is all gone when the block leaves, no? Maybe I am not understanding your problem sorry. The above response is definitely incorrect though (no offense mosquitoman), for one h above the ground is not given, and for another you can't just make up a value for heat because you want it...

the value h above the ground is arbitrary to the start point.. you can pick any points you want.. it wouldnt change delta H

I am not making up the value for heat but calculating it.

 

[Ninjaface]

Assuming the system to be the spring and block together, the total system should not increase or decrease in total energy?

Ah there is the problem. Gravity, an outside force, is doing work on your system. If you want total conservation you are going to have to include the earth in your system. But you don't want that, because this question is about work done on the spring by an outside force (the force acting on the block due to gravity).

Sorry, I'm better at explaining physics stuff to myself than I am to others.

Someone else will likely come along soon and make it clear for you if that didn't...

 

[AZFutureDoc]

I just did:
F=k(dX)
k=F/(dX)
k=10/0.5
k=20

Work done is change in the potential energy, so

W=(dU)=(1/2)(k)(x^2)
W=(1/2)(20)(1/4)
W=2.5

 

[andafoo]

I just did:
F=k(dX)
k=F/(dX)
k=10/0.5
k=20

Work done is change in the potential energy, so

W=(dU)=(1/2)(k)(x^2)
W=(1/2)(20)(1/4)
W=2.5

You say "work done is change in the potential energy", you mean potential energy of the spring? Where does potential energy of the block come into the picture?

And on a side note... how does Work done is the change in kinetic energy (Work-Energy Theorem) come into play? I thought this stuff was easy....

 

[AZFutureDoc]

The energy is converted. If you imaging right when the block is allowed to fall, it has G potential energy. Now, the block falls, and since it's height is now lower, you know that the change in the blocks gravitational energy was transferred to potential energy of the spring.

You converted potential, because the block fell a distance of x. And x is equal to the displacement of the spring. The block lost some potential energy because it is now closer to the surface of the earth. That energy is gained by the spring because it stretches from its eq point (meaning no potential energy) to a displacement of x, (meaning it was given potential energy from the block.)

 

[andafoo]

The energy is converted. If you imaging right when the block is allowed to fall, it has G potential energy. Now, the block falls, and since it's height is now lower, you know that the change in the blocks gravitational energy was transferred to potential energy of the spring.

You converted potential, because the block fell a distance of x. And x is equal to the displacement of the spring. The block lost some potential energy because it is now closer to the surface of the earth. That energy is gained by the spring because it stretches from its eq point (meaning no potential energy) to a displacement of x, (meaning it was given potential energy from the block.)

But why is the quantity of the potential energy lost from gravitational energy not equal to the quantity of the potential energy gained by the spring?

 

[mosquitoman]

andafoo I believe my idea that 1/2 is lost as heat would make sense

I dont see where else that energy could go

 

[andafoo]

andafoo I believe my idea that 1/2 is lost as heat would make sense

I dont see where else that energy could go

I do agree with you. If we have 2 true statements: potential energy of the block is lost and potential energy of spring gains...

Then the only reasonable conclusion is that the energy went to internal energy of the spring... but my problem is that I've never seen that in a text book or problem ever before. It makes me feel like there is something missing... 😕

 

[mosquitoman]

well i thought of a better way to explain it... when you have something that has kinetic energy..and it stops.. that energy either gets converted to potential or gets released as heat.. or both

in our case:

PEbox = PEspring + KEbox

think about it.. the greater the energy of the spring the slower down the box would go (less KE)

so we have

5J = 2.5J + KEbox
so KEbox = 2.5J

but since our box is not moving upwards.. the KE is not converted to PE, but rather to heat since the object comes to a stop

 

[lixxx669]

The energy in fact goes to kinetic energy. If you attach the block on the spring and let go, when the block go to the equilibrium point, it have certein velocity.
But the question seems to assume that the block is forced to keep still in the equilibrium position, which means the energy is not conserve.
But in fact, the key to solve this question is just to know that the definition of elastic potential energy is exactly just the work done by the force of the spring.

 

[mosquitoman]

lix, your explanation is not correct.. the block is not forced to keep still as this would require an outside force.. and we set -kx=mg so there is no outside force

think about it like this.. you take the block and drop it ... it bounces up and down a while and finally comes to rest .5m away

part of energy went to KE and got expelled as heat.. the rest is stored as pe in the spring

 

[lixxx669]

lix, your explanation is not correct.. the block is not forced to keep still as this would require an outside force.. and we set -kx=mg so there is no outside force

think about it like this.. you take the block and drop it ... it bounces up and down a while and finally comes to rest .5m away

part of energy went to KE and got expelled as heat.. the rest is stored as pe in the spring

I agree that if the block is not forced to keep still, then the energy lose would go to the form of heat (friction of air, friction between the molecues of the spring, etc.)
So basically there are 3 situation here:

1. the block is forced to keep still in the equilibrium point --> energy not conserve

2. no outside force, no friction in the system --> the lose of gravitational potential energy turn to kinetic energy + elastic energy (which means at equilibrium point, the block is still moving)

3. the system experienced friction --> Damped Oscillation and finally stop at eq point, the lose of gravitational potential energy turn to heat + elastic energy

But whichever situation it is, the work done by the force of the spring (a conservative force) is only related to k and x, just like that the work done by gravitational force (conservative force too) is only related to G and ^h (the change in h)

 

[andafoo]

I agree that if the block is not forced to keep still, then the energy lose would go to the form of heat (friction of air, friction between the molecues of the spring, etc.)
So basically there are 3 situation here:

1. the block is forced to keep still in the equilibrium point --> energy not conserve

2. no outside force, no friction in the system --> the lose of gravitational potential energy turn to kinetic energy + elastic energy (which means at equilibrium point, the block is still moving)

3. the system experienced friction --> Damped Oscillation and finally stop at eq point, the lose of gravitational potential energy turn to heat + elastic energy

But whichever situation it is, the work done by the force of the spring (a conservative force) is only related to k and x, just like that the work done by gravitational force (conservative force too) is only related to G and ^h (the change in h)

Okay, thanks! I'm glad we've reached a conclusion - it makes sense that the grav. PE is not fully turned into elastic PE (when would that happen? is it even possible?).

 

[BerkReviewTeach]

A vertically oriented spring is stretched by 0.50 m when a mass of 1 kg is suspended from it. What is the work done on the spring?

A. 5.0 J
B. 2.5 J
C. 0.5 J
D. 20.0 J

Correct Answer: B
Explanation


At equilibrium the force of the weight (W = mg) will be equal to the spring force (Fs = -kx), thus:
kx = mg, k = mg/x = (1 kg x 10 m s-2)/(0.5 m) = 20 kg s-2

Work done (spring) = 1/2 kx2 = 1/2 (20)(0.5)2 = 1/2 (20)(1/4) = (20)/(8) = 5/2 = 2.5 J.

My question is, why is the energy stored in the spring not equal to the loss in potential energy of the block?

The work could be found with using F x d, except that proves difficult because the magnitude of the force changes over the pathway (nonconservative force). We could use an average force, but that information was not provided. As a result, we have to solve this using conservation of TOTAL energy.

First, let's assume it starts and finishes motionless, which the question implies. This means that change in kinetic energy is zero.

This means that work = changePE. The trick here is that the system has TWO types of potential energy, spring and gravitational. At different points, posters looked at one or the other, but not both.

As a point of interest (and perhaps reassurance to many who are starting to question themselves), the answer explanation they gave is not the entire picture and I would dare say harmful in that by being incomplete in damages the confidence of test takers. There are often multiple ways to solve questions, and a good answer demonstrtaes that as well as addresses potential issues. But my little diatribe aside, here is what is missing:

The change in potential energy for the entire system must account for BOTH the change in spring and the change in graviational energy. We are assuming the sping to be massless and frictionless, so:

work = mgchangeh - (0.5)kx2

work = (0.5)(1)(10) - (0.5)(20)(0.5)2 = 5 - 2.5 = 2.5

Please note, there is no heat loss in this ideal system. That would be a great deal of heat and hopefully if you've ever stretched a spring you noticed that your fingers didn't get burned.

Also, because the spring potential energy increases as the gravitational potential energy decreases, they are subtracted from one another in the energy equation. The difference represents the work done on the system.

Their solution speaks solely to the spring, which solves their question, but leaves gaps in the overall picture.

 

[mosquitoman]

BerkReviewTeach, I have a question about your explanation

you say that work = change in potential of box - change in potential in spring

this is exactly what i said... change in potential of box = work + change in PE spring

but this work you speak of... where does it go? those 2.5 J... if we were to use conservation the work must be converted either back to potential or released as heat

 

[BerkReviewTeach]

BerkReviewTeach, I have a question about your explanation

you say that work = change in potential of box - change in potential in spring

this is exactly what i said... change in potential of box = work + change in PE spring

but this work you speak of... where does it go? those 2.5 J... if we were to use conservation the work must be converted either back to potential or released as heat

That's where I have issues with their answer. They should have addressed that. The question asks for work done on the spring. The box did work on the spring, so the box experiences -W while the spring experiences +W.

It's similar to when you lift an object. You did work on an object by lifting it. The energy you spend equals the gain in its potential energy. But if we consider the energy of you and the box, then total energy remains constant (assuming you're not solar powered).

 

[andafoo]

That's where I have issues with their answer. They should have addressed that. The question asks for work done on the spring. The box did work on the spring, so the box experiences -W while the spring experiences +W.

It's similar to when you lift an object. You did work on an object by lifting it. The energy you spend equals the gain in its potential energy. But if we consider the energy of you and the box, then total energy remains constant (assuming you're not solar powered).

When it comes to finding Potential Energies and Work it seems like problems always come up in semantics.

The work done by the box will be -W, then. So would there be work done on the box?

The work done on the spring is +W, then what is work done by the spring?

Is there one unifying equation?? It should be the first law of thermodynamics in my opinion... Conservation of Energy. Is there anyway you could wrap it up in a nice package for us?

 

[Kaustikos]

When it comes to finding Potential Energies and Work it seems like problems always come up in semantics.

The work done by the box will be -W, then. So would there be work done on the box?

The work done on the spring is +W, then what is work done by the spring?

Is there one unifying equation?? It should be the first law of thermodynamics in my opinion... Conservation of Energy. Is there anyway you could wrap it up in a nice package for us?

Assigning -/+ to work tells you whether or not work is done on/by the system. If you have -W for something, work is done ON the system or the other way, depending upon how you view this. In the case of the box, it does work on the spring, but not vice-versa - I believe. Berk can correct me on that if incorrect.