What is the motion of the block when released from a compressed spring?

AI Thread Summary
The discussion focuses on the motion of a block released from a compressed spring, detailing its initial acceleration of 160 ft/sec², which decreases linearly as the spring returns to its original length. The equations of motion involve integrating acceleration and velocity to find the time taken for the block to travel specified distances of 4.35 in. and 8.7 in. The user encounters difficulty with the integral of dx/sqrt(320x - 220x²) while attempting to solve for time. The conversation also touches on the potential motion of the block if it were attached to the spring, suggesting oscillatory behavior. Understanding these dynamics is crucial for accurately modeling the block's motion over time.
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Homework Statement



The 15.8-in. spring is compressed to a 7.1-in. length, where it is released from rest and accelerates the sliding block A. The acceleration has an initial value of 160 ft/sec2 and then decreases linearly with the x-movement of the block, reaching zero when the spring regains its original 15.8-in. length. Calculate the time t for the block to go (a) 4.35 in. and (b) 8.7 in.

Homework Equations



a = 160 - kx

vdV = adX
V = dx/dt

The Attempt at a Solution



k = 160/((15.8-7.1)/12)

a = 160 - 220X

Then i integrated vDV = 160 - 220x
and got: v^2 = 320x - 220x^2
solved for V = sqrt(320x - 220x^2)

now I know that V = dx/dt and to solve for dt and integrate for T. However I get stuck at the integral of dx/sqrt(320x - 220x^2) so I assume I am doing something wrong in the process of getting there.
 
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If the block were attached to the end of the spring and the system allowed to continue to move, what kind of motion would you see? What does that suggest for the form of x as a function of t?
 

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