How Does an Imperfect Spring Affect the Velocity of a Released Ball?

AI Thread Summary
The force equation for an imperfect spring is given as F = 150x + 12x^3, where x is the compression in meters. To determine the speed of a 3.0 kg ball released from a 2.0 m compression, the work done by the spring must be calculated using the work-energy theorem. The integral for work should be set from the upper limit of x = 0 to the lower limit of x = 2.0, which reflects the change in potential energy. Concerns about negative values arise from the choice of zero potential, but integrating correctly will yield the necessary positive work done. The discussion emphasizes the importance of defining potential energy reference points when calculating work.
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The force required to compress an imperfect horizontal spring an amount x is given by F = 150x + 12x3, where x is in meters and F in Newtons. If the spring is compressed 2.0m, what speed will it give to a 3.0 kg ball held against it and then released?

I know how to integrate F(x) to get the work done, and with that I could use the work-energy theorem to find the speed. But do I take the lower limit of the integral as x = 2.0 and the upper limit as x = 0?
But then the integral would be negative and 1/2mv2 can't be negative...
 
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Are you finding the work function by solving F = -\frac{dU}{dx}? I would integrate from x=0 to x=x to find the general solution, but if you wanted to put in 2 right away then that would work. The negative thing depends on where you define your zero potential.
 
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