Velocity of Spring: Solved 2m/s

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The discussion revolves around calculating the speed of a block attached to a spring after it returns to its unstretched length. Initially, a force of 40 N is required to hold the block in a stretched position of 40 cm. The spring constant was calculated to be 100 N/m, and the work done by the spring was analyzed using the work-energy principle. The correct approach involves using conservation of energy rather than a constant force, as the spring force varies. Ultimately, the speed of the block when the spring returns to its unstretched length is determined to be 2 m/s.
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[SOLVED] velocity of a spring

Homework Statement


https://wug-s.physics.uiuc.edu/cgi/courses/shell/common/showme.pl?cc/IAState/phys221/spring/homework/05/block_spring_nofriction/4.gif
A spring is stretched a distance of Dx = 40.0 cm beyond its relaxed length. Attached to the end of the spring is an block of mass m = 8.00 kg, which rests on a horizontal frictionless surface. A force of magnitude 40.0 N is required to hold the block at this position. The force is then removed.

When the spring again returns to its unstretched length, what is the speed of the attached object?

Homework Equations



http://wug.physics.uiuc.edu/cc/IAState/Phys221/spring08/course%20info/FS.pdf

The Attempt at a Solution



change in X=.4m
mass=8kg
F=k*dx
calculated k=100N/m
w=f*dl
work=.5mv^2
v=?
projected v=2m/s

How come 2m/s is not working for my speed did I make an error?
 
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Your problem statement is incomplete. What's the question? At what point are you looking for the speed (if that's the question)? Also, calculating work as the given force of 40N times its displacement is not correct. We can assist further once you clarify the problem and note all givens, and have made another attempt at a solution.
 
Sorry, When the spring again returns to its unstretched length, what is the speed of the attached object?
 
I solved it:

Using the spring constant that I found:
F(x) = kx

I found the new Work (W) done for the new distance using:

W = (1/2)k(x2)^2 - (1/2)k(x1)^2

Then with the new amount of work done used the equation Wtotal = (delta)K. So...

W(new) = (1/2)m(v2)^2 - (1/2)m(v1)^2

Starting form it's stretched point the velocity is zero so you're just left with

W(new) = (1/2)m(v2)^2

\sqrt{2}
 
Batman4t said:
Sorry, When the spring again returns to its unstretched length, what is the speed of the attached object?
OK, I see what you may have done, you set the work done by the spring equal to the KE, which is good, but you incorrectly calculated the work. The spring force is variable (it is 0 at the unstretched length), so you can't use a constant 40N as the force...what shouldyou use? Or it is better to use conservation of energy, since only conservative forces are involved..

Edit: OK, you got it.
 
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