Calculating Speed of Mass on a Spring

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A horizontal spring with a stiffness of 0.8 N/m and a relaxed length of 17 cm is stretched to 25 cm with a 21-gram mass attached. When released from rest, the speed of the mass at the relaxed length is calculated using energy conservation principles, where the initial kinetic energy is zero. The relevant equations involve the conversion of potential energy stored in the spring to kinetic energy of the mass. The key to solving the problem lies in correctly identifying the initial and final positions of the spring's compression. The discussion concludes with the poster successfully finding the correct answer after clarifying the formulas and variables involved.
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Homework Statement



A horizontal spring with stiffness 0.8 N/m has a relaxed length of 17 cm (0.17 m). A mass of 21 grams (0.021 kg) is attached and you stretch the spring to a total length of 25 cm (0.25 m). The mass is then released from rest. What is the speed of the mass at the moment when the spring returns to its relaxed length of 17 cm (0.17 m)?

Homework Equations



Ef= Ei +W

Kf + Uf = Ki + Ui + W


The Attempt at a Solution



so no work done W=0
Ki = 0 since released from rest
then 1/2 mvf^2 + 1/2Ksf^2 = 1\2Ksi^2
i want to find vf , i have m , i have k, i guess am having a problem with Sf and Si cause am not getting a correct answer, and am not sure if am using the right formula. Thanks
 
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ok what is the formula when you have a spring ...late me give you hint.

Kf + Uf = Ki + Ui
you don't have to find work because there is no friction .so Uf and Ui is Zero because the hight is 0.so what left

Kf = ki
which is 1/2mv^2 = 1/2kx^2

plug the value you have k,x,and also mass
don't forget there is some trick on X.
good luck .ask if you have question
 
Thanks for your reply , i was able to get the right answer
 
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