# Homework Help: Simple spring-constant equation

1. Jan 10, 2014

### Bindle

1. The problem statement, all variables and given/known data

You decide the spring constant k of a spring by measuring the extension x at different loads. Result is displayed in diagram:

4 cm extension = 10 newton
4.5 cm extension = 11 newton
6 cm extension = 15 newton

The graph which I have is uploaded here: http://nces.ed.gov/nceskids/createagraph/index.asp?ID=5881274468fc4a8e9f509058747271cb (Just press preview)

a) Determine the spring constant.
b) How much work does it take to extend the spring by 4.5 cm?

2. Relevant equations

k = N/d (d = distance)

W = F * d

3. The attempt at a solution

a) Ok, so I start out with deciding the spring constant. I know 4 cm is 10 newton, so I begin:

k = 10 N / 0.04 m
k = 250 N

b) Then I try to figure out the work:

I see that the spring at 4.5 centimeters takes 11 newton of force.

W = 0.045 * 11

W = 0.0495 J

However the solution says it's 0.25 J.

My teacher tried to explain it has to do with the average length, that because of this you have to divide your work in 2. So: W = (0.045*11)/2

W = 0.2475 J and rounded: 0.25 J

Also he tried to explain that I could use the spring constant equation:

k = F/d

k * d = (F/d)*d

k * d * d = F * d

k * d^2 = F * d <- Where I see the point that this is also work.

But still, why do I have to divide the given work in two?

Thanks for any help.

2. Jan 10, 2014

### Staff: Mentor

What is the spring force when the spring is unstretched at x = 0?
What is the spring force when the spring is stretched to x = 4.5 cm?
What is the average spring force during the time that the spring is stretched from x = 0 cm to x = 4.5 cm?
So it doesn't take much work to stretch the spring at the beginning, but it takes much more work to stretch it at the end. That's why you need to use the average to calculate the overall amount of work.

3. Jan 10, 2014

### SteamKing

Staff Emeritus
The units for a spring constant will be N/m or similar.

4. Jan 10, 2014

### rude man

Looks to me like k = k(x). Hooke wouldn't have liked it ...

5. Jan 11, 2014

### Bindle

Oh right, it's 250 N/m. I still don't understand why I have to use the average?

6. Jan 11, 2014

### Staff: Mentor

No really. The force increases in proportion to x (F = kx), but the amount of work needed to increase the length by the same incremental change dx increases as the spring is stretched: dW=kxdx.
Chet

7. Jan 11, 2014

### Bindle

dW = kxdx where dx is delta x or distance times x?

8. Jan 11, 2014

### Staff: Mentor

It's delta x, with delta x being very small. dW is delta W (work) with delta x being very small. For a given Δx, ΔW depends on the amount of x that was already there (i.e., the tension kx that is already in the spring). Every time you try to stretch the spring just a little more, for the same amount of additional stretch, the going gets harder and harder. So you do very little work when you first start to stretch the spring, but you do much more work once the spring already has been somewhat stretched. That's why you use the average force to calculate the total amount of work done.
Chet

9. Jan 11, 2014

### Bindle

Wow, thank you. No I see it.