Results 1 to 3 of 3

Math Help - Work integration

  1. #1
    Newbie
    Joined
    Feb 2009
    Posts
    3

    Work integration

    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
    k=?
    x=4

    Thus, k=13/4.

    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?
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Senior Member mollymcf2009's Avatar
    Joined
    Jan 2009
    From
    Charleston, SC
    Posts
    490
    Awards
    1
    Quote Originally Posted by bottleofboos View Post
    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
    k=?
    x=4

    Thus, k=13/4.

    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.

    Try that first. If it's still wrong, I'll look at the entire integral for you.

    Ok?

    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
    Last edited by mollymcf2009; February 3rd 2009 at 06:15 PM. Reason: Noticed the limits were not right
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Newbie
    Joined
    Feb 2009
    Posts
    3
    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.
    Last edited by bottleofboos; February 3rd 2009 at 08:00 PM. Reason: Didn't see that k also had to be in feet
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Work Integration
    Posted in the Calculus Forum
    Replies: 2
    Last Post: February 7th 2010, 10:39 AM
  2. Integration; work
    Posted in the Calculus Forum
    Replies: 4
    Last Post: March 31st 2009, 04:40 AM
  3. Work and integration
    Posted in the Calculus Forum
    Replies: 2
    Last Post: November 19th 2008, 02:46 PM
  4. Work: Vector Integration
    Posted in the Advanced Applied Math Forum
    Replies: 3
    Last Post: September 18th 2008, 09:07 PM
  5. Amount of work done (integration)
    Posted in the Calculus Forum
    Replies: 4
    Last Post: February 13th 2008, 06:19 AM

Search Tags


/mathhelpforum @mathhelpforum