How Is Work Calculated When Stretching a Spring?

[jumptheair]

Homework Statement

Consider a spring with spring constant k = 8.42 N/m with an unstretched length of 0.250 m. How much work do you do on the spring in stretching it from a length of L1 = 0.450 m to a length of L2 = 0.559 m?

Homework Equations

F= kx
W=(1/2) kx^2

The Attempt at a Solution

since work is area under the curve of x vs. F graph, I did:
W=(1/2)(8.42)(0.559-0.450)^2 = 0.0500 J (wrong answer)
wrong maybe because this works only when u start at the unstretched length?

So I also tried W= (1/2) k xf^2 - (1/2) k xi^2 [where xf = 0.559 and xi=0.450)
=0.464J (still wrong)

What am i doing wrong?

 

[CompuChip]

How do you know it is wrong? Do you have the answer, or are you inputting it in some automated homework system? Because I think you did the right thing (that is, one of your attempts is right) but you didn't round correctly.

The correct attempt is the last one (tell me if you know about integration, then I can tell you why). But also your first remark, about the area under the curve, you are right. Try plotting the curve, marking the area you want to get and then try to see what you did wrong (in particular, also try marking the area you actually calculated in the first attempt).

 

[ogb p]

Your approach to finding the work done on the spring is on the right track, but there are a few things to consider. First, when using the equation W=(1/2) kx^2, the value of x should be the distance that the spring is stretched from its unstretched length. In this case, that would be (0.450-0.250) = 0.200 m. So the correct calculation would be W=(1/2)(8.42)(0.200)^2 = 0.042 J.

Additionally, when using the equation W=(1/2) k xf^2 - (1/2) k xi^2, the values of xi and xf should be the initial and final lengths of the spring, not the distances from the unstretched length. So the correct calculation would be W=(1/2)(8.42)(0.559)^2 - (1/2)(8.42)(0.450)^2 = 0.040 J.

In both cases, the work done on the spring is relatively small, which makes sense since the spring is only being stretched a small amount.