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conservation of energy problem (spring force)



My attempted solution:

Change in KE = Change in spring potential energy

And k = 10^4 using coordinates from the straight line graph

KE1 + PE1 = KE2 + PE2

0.5mv² + 0 = 0 + (- ∫kx)
0.5(20)v² = - 10000 ∫ 20x² (with limits 0.1 and 0)

When I solve this equation I get v = 653m/s.. which is obviously not right. But I don't see the mistake I made. Any help would be appreciated.
Original post by leosco1995

0.5mv² + 0 = 0 + (- ∫kx)
0.5(20)v² = - 10000 ∫ 20x² (with limits 0.1 and 0)


You've confused yourself with this step here- what you've written on the first line is technically valid, but you can't pull k through the integral since it's not a constant in this situation. Then on the second line you seem to substitute x = 20x^2, which is clearly wrong if you think about it.

The much simpler way to solve the problem would be to start with PE = ∫ F dx , (F is of course equal to 20x^2) which is a simple integration.
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leosco1995
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I am a bit confused. If k isn't a constant then what do we even do with it since we can't use it in this equation. I guess we don't need it at all? If I simply do:

KE = PE
0.5(20)v² = 20 * 10³ ∫ x² (with limits 0.1 and 0) (also 20 * 10³ since it's kN)

v turns out to be 0.816 m/s.

The answer is v = 2.38m/s btw.
(edited 9 years ago)
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leosco1995
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I figured it out.. it seems like we need to find the force for the linear deformation and the non-linear deformation too.. and then PE = sum of those. Thanks for the help.
Yeah, the diagram in your question is really badly indicated. The 20x^2 label is referring to the difference between the two lines, so F is actually 10x + 20x^2. If you plug that in your equation from above you get the right answer.

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