Work Problem: Compression of a string

Here is the quesiton:

A spring has a natural length of 2 feet. If a force of 6 lbs is needed to maintain a compression of 6 in., how much work is needed to compress the string from natural length to a length of 16 in.? Round to 4 decimal places.

I'm almost positive I'm solving this correctly but it is wrong according to my online quiz.

F(x) = k * x where k is spring constant and x is distance compressed from natural length.

-6 lb = k * -18 in (Natural length of 2 ft = 24 inches. 24 - 6 in = 18. Negative for compression.)

k = 1/3 lb/in

F(x) = k* x = 1/3x

Integral from 8 to 0 (24 in - 16 in = 8 in) of 1/3x gives me a work value of 10.6667. However this is not the correct answer apparently. I don't understand what I'm doing wrong. I also tried it with converting everything to feet instead of inches and got 3.5556 for final answer, but that is not correct either.

What am I doing wrong? Any help is greatly appreciated

Re: Work Problem: Compression of a string

k isn't right. Look at how the problem is worded.

Re: Work Problem: Compression of a string

Quote:

Originally Posted by

**phys251** k isn't right. Look at how the problem is worded.

I've been staring at this for 10 minutes now and the only other option I see is setting -6 = k * -6. That gives me k equals 1 and the integral from 8 to 0 of (x). The answer still isn't right though. I'm not sure what else to make of it.

Re: Work Problem: Compression of a string

That value of k would be correct if the spring were at rest at 0 feet. Not two.

Re: Work Problem: Compression of a string

Quote:

Originally Posted by

**phys251** That value of k would be correct if the spring were at rest at 0 feet. Not two.

So -6 lb = k * 30 in then? Giving k = 6/30 or 1/5? When I integrate 1/5x from 0 to 8 I get 6.4, which is incorrect as well.

Re: Work Problem: Compression of a string

No. The spring is "centered" at 24 inches. Let x = how far the spring is compressed from that, NOT the distance that the spring is compressed from the 0-inch mark.

So if it's compressed 6 inches, F = 6 as well, giving you your k.