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Work done by spring question

  1. Nov 8, 2006 #1
    Problem statment:

    Trucks have leaf and helper spring (spring put on top of leaf spring vertically) attached to axel. After the leaf spring is compressed by 0.5m, the helper spring helps with additional load.
    k of leaf spring is 5.25*10^5N/m, and k of helper spring is 3.6*10^5N/m. Load is 5.00*10^5N

    a) What is the compression of the leaf spring for the above load?

    I solved by first finding what load leaf spring supports, since its max compression befor helper spring is activated is 0.5m, so i found F with x=0.5, and k=5.25*10^5. Then I subtracted that number from the total load to find how much do both spring support, and then used the sum of both k to find x when two springs are supporting the weight. then 0.5 plus that x is the compression of leaf spring, which gives 0.768m, and thats the answer in the book.

    Now i have a problem with part b:

    b) How much work is done in compressing the springs?

    Any help would be appreciated, thanks!
     
  2. jcsd
  3. Nov 8, 2006 #2

    OlderDan

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    Hint: The work done to compress a spring is the potential energy stored in the spring.
     
  4. Nov 9, 2006 #3
    i did it that way

    took k of 1st spring, times 0.5 squared, and over 2
    then add that
    to k of first and k of second spring, times displacement with 2 springs, 0.268 squared, over 2

    it gives about 0.97*10^5J, which is not the ansewr that book gives...

    book gives 1.68*10^5... i dont know how they get that
    helpp...
     
    Last edited: Nov 9, 2006
  5. Nov 9, 2006 #4
    Each spring has a different displacement. You've actually solved it, you just don't know it yet. The leaf spring is compressed by 0.768 m, the helper spring is compressed by 0.268 m.

    The total work done is the sum of the work done on/by these two springs.

    Dorothy
     
  6. Nov 9, 2006 #5
    ok yeh that worked now...! :)

    i messed up on it because i was adding Ks of both springs for second part of motion, and taking only one spring for first part;, in part a i added Ks so i didnt see why it wouldnt work in part, B, but you are right, its the displacement of each separate spring.

    thanks!
     
  7. Nov 9, 2006 #6
    Ummm. I don't see why that would have worked... I would have solved it this way:

    F = (k_leaf)(x_leaf) + (k_helper)(x_helper)
    x_helper = x_leaf - 0.5m

    And then solved that equation for x_leaf. Are you sure that you added the K's together?

    Well, the right answer is always a good thing to have, in any case.

    Good night,
    Dorothy
     
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