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

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SUMMARY

The motion of a block released from a compressed spring is governed by the equation of motion derived from the spring's acceleration profile. The spring, initially compressed to 7.1 inches, accelerates the block at 160 ft/sec², with acceleration decreasing linearly to zero as the spring returns to its original length of 15.8 inches. The acceleration equation is defined as a = 160 - 220x, where x is the displacement from the spring's original length. The integration of the velocity equation v^2 = 320x - 220x² allows for the calculation of time taken for the block to travel specified distances of 4.35 inches and 8.7 inches.

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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.
 
Physics news on Phys.org
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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