Calculating Speed of Mass on a Spring

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SUMMARY

The discussion focuses on calculating the speed of a mass attached to a spring when it returns to its relaxed length. The spring has a stiffness of 0.8 N/m and a relaxed length of 0.17 m, while the mass is 21 grams (0.021 kg) and is initially stretched to 0.25 m. The relevant equations include the conservation of energy principle, specifically Ef = Ei + W and Kf + Uf = Ki + Ui + W. The correct approach involves setting the kinetic energy equal to the potential energy stored in the spring, leading to the formula 1/2 mvf^2 = 1/2 kx^2, where x is the displacement from the relaxed length.

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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 height 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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