A force of 13 lb is required to hold a spring stretched 4 in. beyond its natural length. How much work W is done in stretching it from its natural length to 6 in. beyond its natural length?
Using Hooke's Law, f(x)=kx , I set it up like this:
f(x) = 13
I plug that in to the to integral of 13x on the interval from 4 to 6. Then I got 65/2 in-lbs. I divide by 12 becuase the problem requires it to be ft-lb. Thus I got I got 65/24 ft-lbs, but the site says it is wrong. What did I do wrong?
First thing I noticed is that the limits on your integral are in the wrong units and the lower limit is wrong. If your problem is in ft-lbs, your limits must also be in ft-lb and as it looks, they seem to be still in inches.
Originally Posted by bottleofboos
Try that first. If it's still wrong, I'll look at the entire integral for you.
One other thing, your bottom limit should start at 0, or natural length of the spring. Just because you use the force given at 4in to calculate your k, does not mean that it automatically includes that in your answer. You must evaluate the entire interval from 0 to 6 inches (or in this ft-lb situation) 0 to .5ft
Okay, so I tried it on the interval of 13x/4 from 0 to 1/2 ft. So i got 13/32 ft-lbs.
I wanted to double check using the the interval from 0 to 6 in and then divide the answer by 12 to convert it into ft-lbs. So i got 39/8 ft-lbs.
Which is the right answer?
EDIT: I found out that the k has to be converted into feet also if I want to use the interval from 0 to 1/2 ft. So now I got 39/8 ft-lbs for each side. Thanks Molly.