Find Energy Stored in Two Parallel Springs

AI Thread Summary
The discussion revolves around calculating the initial velocity of a block compressing two parallel springs, one inside the other, using energy conservation principles. The user successfully derived the equation for velocity using the energy stored in the springs, applying Hooke's Law and integrating the force over distance. A misunderstanding regarding the distance compression was clarified, suggesting that the distance should be considered as d1 + d2 rather than just d2. The user confirmed that integrating the force vs. distance curve provided a consistent result with the energy stored in each spring. The final equation aligns with the energy contributions from both springs, confirming the correctness of the approach.
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Homework Statement


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.


Homework Equations


F = -kx
U = 1/2 mv^2 = 1/2 kx^2


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!
 

Attachments

  • Two Springs Setup.JPG
    Two Springs Setup.JPG
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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.
 
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:

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

\ \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)

\ \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

\ \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

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