Six and one-half foot-pounds of work is required to compress a spring 4 inches from its natural length. Find the work required to compress the spring an additional one-half inch. (Round your answer to two decimal places.)

This is what I did so far:

4 in = 1/3 ft

6.5 = ∫[0,1/3] kx dx
6.5 = k(x^2/2)(Eval at 1/3 and 0)
6.5 = k/18
117 = k

Now figuring the Work:
W = ∫[1/3,5/6] 117x dx
W = 117(x^2/2) (Eval at 5/6 and 1/3)
W = 34.125 ft-lb (34.13 ft-lb) but Webassign is counting that wrong. I re-looked at my problem and everything but I can't seem to find what's wrong with my work. Help please?

Well, it seems like you're having a spring-loaded problem with your calculations! Let me give it a shot and see if I can bring some humor to your predicament.

First of all, kudos to you for doing all that math! You really spring into action when it comes to problem-solving. But it seems like you took a little hop in the wrong direction at the end.

When calculating the work required to compress the spring an additional one-half inch, you made a small leap in your integral limits. Instead of integrating from 1/3 to 5/6, you should integrate from 1/3 to 1/3 + 1/2. That's right, we want to find the work required to compress the spring an additional one-half inch from its current position!

So let's recalculate the work, shall we?

W = ∫[1/3, 1/3 + 1/2] 117x dx

And after integrating, we get:

W = 117(x^2/2) (Eval at 1/3 + 1/2 and 1/3)
W = 117((5/6)^2/2 - (1/3)^2/2)

Now, let's plug in those numbers and see what we get:

W = 117((25/36)/2 - (1/9)/2)
W = 117((25/72) - (1/18))
W = 117(23/72)
W ≈ 37.79 foot-pounds

And there you have it! The work required to compress the spring an additional one-half inch is approximately 37.79 foot-pounds. So go ahead and spring that answer on Webassign, and hopefully, this time, it will come out laughing!

Your calculation seems correct, and the answer you obtained, 34.13 ft-lb, is the correct answer. It's possible that there might be a rounding issue or a different formatting requirement for your answer in Webassign. Double-check if there are any specific instructions for expressing your answer, such as rounding to a specific decimal place or using a different unit of measurement. If you are confident in your calculations, you can consider reaching out to your instructor or TA for clarification on any grading discrepancies.

To find the work required to compress the spring an additional one-half inch, let's break down the problem step by step.

Step 1: Find the value of the spring constant (k).
We are given that 6.5 foot-pounds of work is required to compress the spring 4 inches. We can use this information to find the value of k.

We know that work (W) done on a spring is given by W = (1/2)kx^2, where k is the spring constant and x is the displacement.

So, we have 6.5 = (1/2)k(4^2).
Simplifying, we get 6.5 = 8k.
Dividing both sides by 8, we find k = 0.8125.

Step 2: Find the work required to compress the spring an additional one-half inch.
Now that we have the spring constant (k), we can calculate the work required to compress the spring an additional one-half inch (0.5 inches).

We need to calculate the work done on the spring from the initial position (x = 4 inches) to the final position (x = 4 + 0.5 = 4.5 inches).

The work done on the spring (W) is given by W = (1/2)k [(4.5)^2 - (4)^2].
Simplifying, we get W = 0.5 * 0.8125 * (4.5^2 - 4^2).
Calculating this expression, we find W ≈ 0.708 ft-lb (rounded to two decimal places).

So, the work required to compress the spring an additional one-half inch is approximately 0.708 ft-lb.

Woops, I didn't convert 1/2 inch to feet. the W equation should be [1/3,9/24], leading to the answer to be around 1/73 ft-lb