Solving Spring Stretching Problems: Calculating Work

[thename1000]

Homework Statement

For example:

1.) A spring has natural length 1 m. A force of 24 N stretches it to a length of 1.8 m.
How much work is required to stretch the spring 2 meters beyond its natural length? (ans 60 J)

2.) Suppose that 2J of work is needed to strech a spring from its natural length of 30 cm to a length of 42 cm.
How much work is needed to stretch it from 35 cm to 40 cm.

The Attempt at a Solution

Generally in these problems you solve for K, and then get the integral of kx. how do I set up the limits of these two integrals? In my notes for #2, I have two integrals. the first is used to find K. Is this done always? Is this done for #2? If so what are the limits?

It seems it should be integrating from amount of first strech to total stretch. but I'm not sure.

Any help?

thanks

 

[Galadirith]

Hi thename1000,

So you've solved 1.). Now for two I think the best way is to have two integrals as you have suggested. I would say yes this is done always, but then this is quite a specific question, there's really only this way to solve it.

So required equation:

<br /> E_{el} = \int_{x_0}^{x_1} kx \ dx \ = \ \frac{1}{2}kx^2 \Bigr|_{x_0}^{x_1} \ = \ \frac{1}{2}k\left( x_1^2 - x_0^2\right)<br />

So breaking up question 2 into two part. First part is as you have already realized yourself, to find k. You know that the work required is 2J (E_el) and you do know the limits, given that it is stretched from 30cm to 42cm and its natural length is 30cm. Obviously it is being stretched from its natural length therefore x₀ =0.

For the second part your probably unsure as to what the limits should be because it is not being stretched from its natural length? But it is very simple, it is being stretched from 35cm to 40cm so from and extension of x₀, where x₀ ≠ 0, to x₁.

This is what you suggested in you final comment and it is exactly correct. Hopefully that's helped thename1000 :D

 

[thename1000]

Hey Galadirith, thanks for the response.

I know the answer for #1, but not how to do it, do you think you could walk me through how to do that one? Do I need to integrate to find K, or can I just do W=k*amnt stretched.

Thanks!

 

[Galadirith]

Oh ok thename1000, sure think I just assumed since the answer was there.

So question #1, you need to first find k. Now you don't need to integrate anything to find k for the first question as we are given the force and the length of the spring from which we can determine the extension, so we can use F = kx = k*(amount stretched from natural length correctly know as the extension), I assume W means weight in your equation and not work :D.

Now be careful with how you interpret the equation F = kx, because technically that's the correct form of the equation, it should be:

<br /> F = -kx<br />

where F is tension in the spring, and x is the extension in the spring. Now it also has a negative value, because the tension force acts in the opposite direction to the direction of extension. You may already know this but its good to reiterate :D

The reason why we can write it as F=kx and not F=-kx is all to do with whether F represents the tension in the spring (generally the elastic material) or it represents the force extending the spring, caused either by someone who has pulled it or a mass suspended on the end of the spring.

Anyway back to the question. So we have found k now we need to find the work done to extend the spring to 2m. Now another way of looking at that is how much Elastic Potential energy is there stored in the spring. Now hopefully you have been told how one can find this energy, as understand why is important, but the long and short of it is we integrate hooks law (the F=kx) with respect to x, thus arriving at the equations that I previously posted.

Now bar actually telling you how to do this problem I think that's all I can say. Have a go at doing 1.) yourself and 2.) if you can, then once you have done as much as you can if you can't get the answer post back here and well try to give you gentle nudges in the right direction :D