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Find Energy Stored in Two Parallel Springs

  1. Feb 3, 2010 #1
    1. The problem statement, all variables and given/known data
    There are two springs, one long spring of constant k1 and it is inside a smaller spring of constant k2. Both springs follow Hooke's Law. A box of mass M moves without friction and compresses spring 1 a distance d1 where it hits spring 2. Spring 1 and 2 then compress further distance d2. Solve for the initial velocity of the block.

    2. Relevant equations
    F = -kx
    U = 1/2 mv^2 = 1/2 kx^2

    3. The attempt at a solution
    I was able to get the solution from the following: (1/2)M v^2 = 1/2 k1 d2^2 + 1/2 k2 (d2-d1)^2 and solving for v. I also thought I could get the solution from finding the area under a force vs distance curve but I was unsuccessful. In this attempt I found the area under the small triangle formed by k1 slope, the area of the larger triangle formed by k1+k2 and the little rectangle beneath.

    The equation was: 1/2 k1 d1^2 + 1/2 [(k1+k2)d2 - k1d1]*(d2-d1) + k1d1d2

    I attached a picture of the initial scenario and graph. Any help is greatly appreciated, thanks!

    Attached Files:

  2. jcsd
  3. Feb 3, 2010 #2
    It says a further distance d2 which you have interpreted as d2-d1.
    This may be the problem. Try taking it d1+d2 on graph instead of d2 and check for the answer.
  4. Feb 6, 2010 #3
    I wanted to update- I found the area under the curve by integrating and it matched the answer as if we considered the energy stored in the compression of the springs separately:

    [tex]\int ^{d1}_{0}k_{1}x +\int ^{d2}_{d1} (k_{1}+k_{2})x +k_{1}d_1[/tex]

    [tex]\ \frac{1}{2} k_{1}d_{1}^2 + \frac{1}{2}(k_1+k_2)(d_2-d_1)^2 +k_1d_1(d_2-d_1)[/tex]

    [tex]\ \frac{1}{2}k_1d_1^2 + \frac{1}{2}(k_1d_2^2 +k_1d_1^2-2k_1d_1d_2 +k_2d_2^2+k_2d_1^2 - 2k_2d_1d_2) +k_1d_1d_2 - k_1d_1^2 [/tex]

    [tex]\ \frac{1}{2}k_1d_2^2 + \frac{1}{2}k_2d_2^2 + \frac{1}{2}k_2d_1^2 - k_2d_1d_2 [/tex]

    And this is the same as the two springs separately:
    [tex]\ \frac{1}{2}k_1d_2^2 + \frac{1}{2}k_2(d_2-d_1)^2 [/tex]
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