How Much Work to Stretch a Spring 4 cm from Equilibrium?

[flyingpig]

Homework Statement

When a 5kg mass is hung vertically on a light spring the spring stretches 3 cm. how much work must a external agent do to stretch the spring 4 cm from its equilibrium position?

Because I am guessing it is hung vertically from the ceiling?

 

[fluidistic]

Homework Statement

When a 5kg mass is hung vertically on a light spring the spring stretches 3 cm. how much work must a external agent do to stretch the spring 4 cm from its equilibrium position?

Because I am guessing it is hung vertically from the ceiling?

No you don't need the height.
Yes it's hung vertically from the ceiling. I'll give you a hint to solve the problem: consider the potential energy stored in a string.

 

[flyingpig]

Why don't you need to know the height? Isn't there gravity? I know that it is just

\int_{x_0 = 3cm}^{x = 4cm} -kx dx

But I am sure that I need gravity

 

[Fewmet]

Why don't you need to know the height? Isn't there gravity? I know that it is just

\int_{x_0 = 3cm}^{x = 4cm} -kx dx

But I am sure that I need gravity

You need the acceleration due to gravity, with varies do little with height that it is treated as a constant 9.81 m/s² near the surface of the Earth.

Do you see how to calculate k from the given information?

Also note that the question asks for the work done stretching from the equilibrium position to 4 cm. The integral you posted is inconsistent with that.

 

[fluidistic]

You need the acceleration due to gravity, with varies do little with height that it is treated as a constant 9.81 m/s² near the surface of the Earth.

Do you see how to calculate k from the given information?

Also note that the question asks for the work done stretching from the equilibrium position to 4 cm. The integral you posted is inconsistent with that.

This is right.
Although I do find the wording of the question confusing. When they say

how much work must a external agent do to stretch the spring 4 cm from its equilibrium position?

, do they mean equilibrium position when there's the mass attached to the spring or when the mass isn't attached yet? Because if it is the former then flyingpig is right, I believe.

 

[flyingpig]

[PLAIN]http://img824.imageshack.us/img824/403/hieh.jpg

I need to know height, this is a picture of how I interpret the problem

 

[Sourabh N]

Let me rephrase the question. This might make it clear what the question is asking.

How much work must a external agent do to stretch a light spring 4 cm from its equilibrium position? Oh by the way, when a 5kg mass is hung vertically on the spring the spring stretches 3 cm.

 

[Fewmet]

Let me rephrase the question. This might make it clear what the question is asking.

How much work must a external agent do to stretch a light spring 4 cm from its equilibrium position? Oh by the way, when a 5kg mass is hung vertically on the spring the spring stretches 3 cm.

That's how I was seeing it. The "by the way" clause let's you find k, and that then gets you an answer.

 

[fluidistic]

Let me rephrase the question. This might make it clear what the question is asking.

How much work must a external agent do to stretch a light spring 4 cm from its equilibrium position? Oh by the way, when a 5kg mass is hung vertically on the spring the spring stretches 3 cm.

Ohhh... I see now. Personally I thought flyingpig had its limits of the integral right. So thanks a lot for the clarification.
And who else agree with me when I say that flyingpig doesn't need the height? As long as g is treated as a constant, there's no need to know the height.

 

[flyingpig]

What...?

 

[Sourabh N]

What...?

Read the posts after yours and point out the line which you do not understand.

 

[flyingpig]

Read the posts after yours and point out the line which you do not understand.

Let me rephrase the question. This might make it clear what the question is asking.

How much work must a external agent do to stretch a light spring 4 cm from its equilibrium position? Oh by the way, when a 5kg mass is hung vertically on the spring the spring stretches 3 cm.

Isn't that what I did with my integral? BUt my integral is still only half completed.

 

[Fewmet]

Isn't that what I did with my integral? But my integral is still only half completed.

I see I was mistaken about the integral. At a glance I thought you were using it to find the work, but now I see that expression is intended to equal force. IN that case the limits of integration are correct.

 

[PieOperator]

Homework Statement

When a 5kg mass is hung vertically on a light spring the spring stretches 3 cm. how much work must a external agent do to stretch the spring 4 cm from its equilibrium position?

Because I am guessing it is hung vertically from the ceiling?

Although springs exert a nonconservative force, you can treat this problem as a conservative force problem for now. Conservative forces have a definite integral and a potential energy graph. Thus, you need to know the height when using the ever easy equation of potential energy equals mass times gravity times height. Just think about how you can apply the 2nd law of thermodynamics into this great and insightful setting!

 

[fluidistic]

Although springs exert a nonconservative force, you can treat this problem as a conservative force problem for now. Conservative forces have a definite integral and a potential energy graph. Thus, you need to know the height when using the ever easy equation of potential energy equals mass times gravity times height. Just think about how you can apply the 2nd law of thermodynamics into this great and insightful setting!

Look at post #6. When you say "you need to know the height when using the ever easy equation of potential energy equals mass times gravity times height.", flyingpig might think he needs to know the value of d or h. And he doesn't need these values.

 

[flyingpig]

Okkay, how about this, tell me why don't you need the height? Why is there no gravity in play? It's hanging vertically

 

[Fewmet]

Okkay, how about this, tell me why don't you need the height? Why is there no gravity in play? It's hanging vertically

I suspect you are thinking about this differently than some of us, and the confusion is coming from that. Can you say how you would use the height if it were given? That should let us explain why it is not necessary.

 

[flyingpig]

[PLAIN]http://img200.imageshack.us/img200/92/unledvx.jpg

\sum W = \int_{x_0 = 3\cdot 10^{-2}m}^{x = 4\cdot 10^{-2}m} -kx dx + -mg(5m - 6m)

 

[Fewmet]

\sum W = \int_{x_0 = 3\cdot 10^{-2}m}^{x = 4\cdot 10^{-2}m} -kx dx + -mg(5m - 6m)

Thanks for clarifying your approach.

Here is the original question:

When a 5kg mass is hung vertically on a light spring the spring stretches 3 cm. how much work must a external agent do to stretch the spring 4 cm from its equilibrium position?

The work done by the "outside agent" depends on the force he/she/it applies to the 5 kg mass. The -kx is the upward force exerted by the spring. The mg is the force exerted by gravity. Where is the force exerted by the outside agent?

 

[fluidistic]

[PLAIN]http://img200.imageshack.us/img200/92/unledvx.jpg

\sum W = \int_{x_0 = 3\cdot 10^{-2}m}^{x = 4\cdot 10^{-2}m} -kx dx + -mg(5m - 6m)

I'm afraid this sketch is wrong. Seems like you mean that the weight of the mass stretched the spring 1 meter while this isn't true.
Also we can see that 1cm=1m from the sketch.

The work done by the "outside agent" depends on the force he/she/it applies to the 5 kg mass. The -kx is the upward force exerted by the spring. The mg is the force exerted by gravity. Where is the force exerted by the outside agent?

So now we interpret the problem as stretching the spring 4 cm from its equilibrium position (when there's no mass attached) when a mass of 5 kg is hanging from the spring?

 

[Fewmet]

So now we interpret the problem as stretching the spring 4 cm from its equilibrium position (when there's no mass attached) when a mass of 5 kg is hanging from the spring?

That's how I am reading it, fluidistic, although I wonder if flyingpig rephrased a little when transcribing it to the original post.

flyingpig: Something further occurred to me as I thought more about this. You supposed the heights were 5 m and 6 m (although fluidistic is right that the stretch is only 1 cm). Do you see that no matter what heights you choose the ∆x will still be 1 cm? That's why you don;t need to know the height, just the change in height.

 

[fluidistic]

Thanks Fewmet.
Now hopefully flyinpig will understand why he doesn't need the height.

 

[flyingpig]

OKay sorry I meant cm,

So let's just pretend the picture had 6m = 6cm and 5m = 5cm.

Fewmet said we only need the change in height, but fluid said we don't need it at all??

 

[Fewmet]

Fewmet said we only need the change in height, but fluid said we don't need it at all??

You don't need the height in that whether this was 1 m off the follow or 50 m, the answer will still be the same. You just need that the mass is getting 1 cm farther from the equilibrium position.

So you want to know how much work the "outside agent" does in pulling the mass that extra 1 cm. That means you need an expression for the the force applied that did that work.

You are really in the verge of getting an answer.

I will be away from the computer for the next 6 hours or so, so I'll not be posting replies anytime soon.

 

[omega_minus]

Flyingpig, have you tried a free body diagram yet?

 

[Libohove90]

Homework Statement

When a 5kg mass is hung vertically on a light spring the spring stretches 3 cm. how much work must a external agent do to stretch the spring 4 cm from its equilibrium position?

Because I am guessing it is hung vertically from the ceiling?

Simply put... Net work = 1/2 kx^2 - mgy where x is the displacement from the initial equilibrium position to the final position and y is the vertical displacement due to gravity. You see that the work done on the mass by gravity is mgy...thus gravity helps you. Thus subtracting the total work done in order to pull this mass to the final position with the work done by gravity gives you the work you need to stretch it an extra cm. You don't need height, you just need the vertical displacement due to gravity which is mgy. Height is irrelevant here because it measures the distance from the ground, but we don't need that here because 1) we know the vertical displacement due to gravity and 2) even if we did know the height, vertical displacement would be the same.

 

[fluidistic]

OKay sorry I meant cm,

So let's just pretend the picture had 6m = 6cm and 5m = 5cm.

Fewmet said we only need the change in height, but fluid said we don't need it at all??

We need the change in height. We don't need the height at all.
Actually your ceiling could be 1 m over the ground or 100 m, it doesn't change anything and you don't need these values. The change in height wouldn't change (as long as g doesn't change).

 

[omega_minus]

Libohove90, when I evaluate your expression, I get a negative answer, even though the displacement vector and force vector are parallel (and thus the work done by the agent is positive). I simply calculated k with the same method you did, then evaluated the integral from 0.03m to 0.04m.

 

[flyingpig]

Do I even need gravity in this question or not??

 

[Fewmet]

Do I even need gravity in this question or not??

You should be able to answer that question on your own by setting up the equations and seeing what is in them. Doing that will show you that you need the acceleration due to gravity (which you have already used several times). You also need the weight of the 5 kg mass (which is the force of gravity acting on the mass), but that is the mg=49 N.

 

[omega_minus]

From the sound of your question it sounds like you're still working the problem from the floor up. It's not a gravitational potential energy problem, as others have said. Work the problem from the ceiling down, from where the spring is hung. Those are the measurements that matter. The height above the floor is not important, but as Fewmet has said g will play a role in the calculation.

 

[flyingpig]

I see it now, I overlooked it.

\sum W = \int_{x_0 = 3.00cm}^{x = 4.00cm} -kx\cdot dx - mg(-1.00cm)

Right? I have negative change in distance because I am climbing "down hill"

 

[ideasrule]

If you compute that answer, you'll find that it's negative. That's because you calculated the work exerted by the spring on the external agent, not the work exerted by the external agent on the spring.

Here's a useful way to approach this problem. We're applying a force to pull the spring down, and that force is kx. Gravity provides some help, so we only really need to apply kx-mg. The integral of kx-mg from x=3cm to x=4cm is the work that the external agent needs to do.

 

[flyingpig]

Well we don't even know what k is

I calculated and I got

-3.5 * 10^-4 * k + 50 * 1cm= Work

 

[ideasrule]

The question tells you that when a 5kg mass is hung on the spring, it stretches by 3cm. A 5kg mass applies a force of (5 kg)*(9.8 m/s^2), so can you find the spring constant?

 

[flyingpig]

Oh lol

mg = kx

mg/x = (5*10)(3) = 50/3 = 16.7N/m

16.7N/m * -3.5 * 10^-4 + 50 * 1 * 10^-2 = 0.49J Still positive?