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Work done by a Spring and such

  1. Mar 11, 2008 #1
    Hey guys, this is an example from a quiz last semester that im studying off of. However I cant figure out how to solve this question. Here is goes:

    1. Block A, with Mass m, sits on top of a spring which is compressed 10 cm. When the spring is released, the block is launched to a maximum height of 2m above its starting point. Block B, with mass 2m, is placed on the same spring and launched 2m above the starting point.

    a) how far should the same spring be compressed to launch block A 4m above its starting point?

    b) how far should the same spring be compressed to launch block B to a max hight of 2m above its starting point.


    For part A, I ended up with 40 cm. However Im pretty sure thats wrong. if someone can take me through the thought process and help id really appreciate it.
     
  2. jcsd
  3. Mar 11, 2008 #2
    Work by a spring and such

    1. The problem statement, all variables and given/known data

    Hey guys, this is an example from a quiz last semester that im studying off of. However I cant figure out how to solve this question. Here is goes:

    1. Block A, with Mass m, sits on top of a spring which is compressed 10 cm. When the spring is released, the block is launched to a maximum height of 2m above its starting point. Block B, with mass 2m, is placed on the same spring and launched 2m above the starting point.

    a) how far should the same spring be compressed to launch block A 4m above its starting point?

    b) how far should the same spring be compressed to launch block B to a max hight of 2m above its starting point.




    2. Relevant equations

    hookes law: 1/2kxf^2 + 1/2kxi^2 = W
    Kinetic Energy: W = 1/2mv^2
    Kd= mg
    * I may need a different equation im not aware of


    3. The attempt at a solution

    Part A)
    I tried solving for the compressiob of the spring by setting up 2 equations. Kd=mg, i solved for K to get K = m(9.8)/.1m

    Then I tried solving for x, using the equation ws + wg = 0 , so 1/2Kx^2 -mgh = 0 --> x = sqrt(2(mgh/k) ---> x = sqrt(2(m(9.8)(2m)/m(9.8)/.1m))) and got 40 cm as the resulting compressed spring.

    Im pretty sure that its wrong, I am just overwhelmed and confused
     
  4. Mar 11, 2008 #3

    Doc Al

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    Staff: Mentor

    Why don't you explain how you ended up with 40 cm. Hint: Energy is conserved.
     
  5. Mar 11, 2008 #4
    Well, heres what I did..although it could be way off.

    I set kd=mg, and solved for k. the answer i got was k = m(9.8)/.1m.
    Then i used the equation ws + wg = 0 => 1/2Kx^2 - mgh = 0 => x = sqrt((2mgh)/k) then i plugged in the numbers and ended up with 40 cm.

    but I think d is equal to 2.1 and not .1, correct?

    *this whole thing may be wrong, but thats as far as I can get by myself.
     
  6. Mar 11, 2008 #5

    Doc Al

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    Staff: Mentor

    You're not given any data to solve for k, but you don't need to solve for k to answer the questions. (Just because the spring is compressed 10 cm, does not mean that the weight alone is doing the compressing.) Just call the spring constant "k".

    What you need to do is track energy changes: from spring PE to gravitational PE.

    Hint for part a): By what factor must the energy change for mass A to reach a height of 4 m?
     
  7. Mar 11, 2008 #6
    alright, I figured out part A. I used the PEs = PEgrav equation, which is 1/2Kx^2 = mgh
    I plugged in .1 for x for part A (along with all of the other given info) and the answer i came up with was 3920m = K, and I plugged it into the equation again to get x = .141m as the amount its compressed.

    Now my question is, for part B, is the "x" value still .1? because it doesnt specify.
     
  8. Mar 12, 2008 #7

    Doc Al

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    Staff: Mentor

    No you cannot assume that the compression remains the same, since that's what they are asking you to find. The spring constant is the same though, since its the same spring.

    Hint: How much energy is needed to launch block B to the height specified?
     
  9. Mar 12, 2008 #8
    hmm, well if the spring constant is the same, i got the same answer for part B as I did for A. because B is twice the mass, but goes half as high. Is that correct? or am I still missing something
     
  10. Mar 12, 2008 #9

    Doc Al

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    Staff: Mentor

    No, not correct. Block B is launched 2m. You know how much spring compression is needed to launch block A 2m; so how much is needed to launch block B (which has twice the mass) the same height?

    How much energy is needed? How does spring energy depend on compression?
     
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