Finding velocity when a mass suspended from a spring passes through equilibrium

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SUMMARY

A 1kg mass suspended from a spring with a spring constant of 100 N/m is compressed by 10 cm before being released. The potential energy stored in the spring (PEelastic) is calculated using the formula PEelastic = 1/2 kx^2, resulting in 5 Joules. At the equilibrium position, this potential energy converts entirely into kinetic energy (KE), allowing the velocity of the mass to be determined using KE = 1/2 mv^2. The final velocity of the mass as it passes through the equilibrium position is 3.16 m/s.

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A 1kg mass is suspended from a vertical spring with a spring constant of 100 N/m and the equilibrium position is noted. The spring is then pushed upward (compressed) a distance x=10 cm before the mass is released from rest. How fast will the block be moving when it passes through the equilibrium position?


PEelastic=1/2kx^2
KE=1/2mv^2




I honestly have no idea what I'm doing. I used the equation above and got: 1/2(100)(10)^2 which equals 5000, but I'm honestly not sure what to do after I get the PEelastic (or if I'm even supposed to solve for that). I think if I could figure out KE, I could solve for velocity, but I'm not sure how to solve for kinetic energy...any help in the right direction would be great...
 
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You are going in the right direction by finding PEelastic. By doing this you also found KE since KE=PEelastic so from there you just plug it into the equation.
 

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