Advertisement
If you have a new account but are having problems posting or verifying your account, please email us on hello@boards.ie for help. Thanks :)
Hello all! Please ensure that you are posting a new thread or question in the appropriate forum. The Feedback forum is overwhelmed with questions that are having to be moved elsewhere. If you need help to verify your account contact hello@boards.ie
Hi there,
There is an issue with role permissions that is being worked on at the moment.
If you are having trouble with access or permissions on regional forums please post here to get access: https://www.boards.ie/discussion/2058365403/you-do-not-have-permission-for-that#latest

calculation of force through calculus

  • 01-03-2014 11:27am
    #1
    itgirl268
    Registered Users, Registered Users 2 Posts: 55 ✭✭


    Work of 6 Joules is done in stretching a spring from its natural length to 21 cm beyond its natural length. What is the force (in Newtons) that holds the spring stretched at the same distance (21 cm)?
    iv tried it several times but answer is incorrect..? think im calculating it incorrectly


Comments

  • #2
    TheBody
    Registered Users, Registered Users 2 Posts: 5,633 ✭✭✭TheBody


    itgirl268 wrote: »
    Work of 6 Joules is done in stretching a spring from its natural length to 21 cm beyond its natural length. What is the force (in Newtons) that holds the spring stretched at the same distance (21 cm)?
    iv tried it several times but answer is incorrect..? think im calculating it incorrectly

    It's been a LONG time since I've done any physics, but from my dodgy memory:

    Work=Force x Distance

    Rearranging this tells us that:

    [latex]Force=\frac{Work}{Distance} [/latex]

    From that, can you figure out the answer?


  • #3
    itgirl268
    Registered Users, Registered Users 2 Posts: 55 ✭✭itgirl268


    so i done it 6J/.21 = 28.57 but this is incorrect am i leaving out something??


  • #4
    TheBody
    Registered Users, Registered Users 2 Posts: 5,633 ✭✭✭TheBody


    I'm afraid, I have to bow out to superior minds on this.

    I'll stick to doing maths!!


  • #5
    bnt
    Registered Users, Registered Users 2 Posts: 13,077 ✭✭✭✭bnt


    Another way of looking at it: the Work = Force x Distance formula assumes the force is constant but, in the case of a spring, it's not. It's proportional to the extension/compression of the spring. (F = k x, where k is the spring constant).

    So, you're not just multiplying force by distance, you're integrating the force function over distance:

    W = ∫ F dx
    = ∫ k x dx
    = ½ k x²

    From that you can calculate k, then the force.
    k = 272,1, F = 57.14

    You are the type of what the age is searching for, and what it is afraid it has found. I am so glad that you have never done anything, never carved a statue, or painted a picture, or produced anything outside of yourself! Life has been your art. You have set yourself to music. Your days are your sonnets.

    ―Oscar Wilde predicting Social Media, in The Picture of Dorian Gray



  • #6
    itgirl268
    Registered Users, Registered Users 2 Posts: 55 ✭✭itgirl268


    would a similar idea be applied to this as above????

    A chain 67 meters long whose mass is 27 kilograms is hanging over the edge of a tall building and does not touch the ground. How much work is required to lift the top 7 meters of the chain to the top of the building? Use that the acceleration due to gravity is 9.8 meters per second squared.


  • Advertisement
  • #7
    dlouth15
    Registered Users, Registered Users 2 Posts: 1,169 ✭✭✭dlouth15


    itgirl268 wrote: »
    would a similar idea be applied to this as above????

    A chain 67 meters long whose mass is 27 kilograms is hanging over the edge of a tall building and does not touch the ground. How much work is required to lift the top 7 meters of the chain to the top of the building? Use that the acceleration due to gravity is 9.8 meters per second squared.
    A similar approach is required here because you have a force that varies as the chain is being drawn in. You need to express that force as a function of the length of chain still hanging. You then need to derive an expression for the small bit of work required to haul the chain in by an infinitesimal length. The total work will be the sum of all these small bits which will involve an integral.


Advertisement