Mass on a vertical Spring: finding maximum speed

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The discussion focuses on a physics problem involving a mass on a vertical spring with a spring constant of 60 N/m. The first part of the problem determines that the spring stretches 0.245 m when a 1.5 kg block is hung from it. In the second part, the block is pulled down to stretch the spring an additional 0.245 m, making the total stretch 0.49 m, and is then given an initial speed of 3.7 m/s. The user initially miscalculated the maximum speed of the block, arriving at an incorrect value of 4.82 m/s, due to neglecting the work done by gravity. The correct approach involves treating the stretched position as the new equilibrium and recalculating using the appropriate energy conservation equation.
ilyketurdles
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Homework Statement



A spring with spring constant k = 60 N/m and unstretched length of L0 = 2.2 is attached to the ceiling. A block of mass m = 1.5 kg is hung gently on the end of the spring.
1)

How far does the spring stretch?

Answer is .245 m

2)

Now the block is pulled down until the total amount the spring is stretched is twice the amount found in the above question. The block is then pushed upward with an initial speed vi = 3.7 m/sec. What is the maximum speed of the block?

Homework Equations



F spring = -kx
Wnc = change in ME[/B]

The Attempt at a Solution



I found the first part just by using force, with the force required by the spring to be equal to mg, then found x from there.

For part 2, however, I'm having problems.

I took x to be .245 m (from the first part) x 2 = .49 m

Then, assuming the forces are all conservative, and that the velocity will be maximum at the spring's relaxed position, I used the formula:

.5 mv02 + .5kx2 = .5 mvf2

Solving for Vf, I get 4.82, which is incorrect. I've also tried doubling the entire spring, in case that's what the question was asking, and was still incorrect.
 
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So I figured it out, leaving my method of solution for anyone else who may come across this problem.

I neglected work due to gravity. So what you can do is take the stretched position in problem 1 as the equilibrium position, then use the formula I used before

.5 mv02 + .5kx2 = .5 mvf2

except x is now the amount stretched for the already stretched position, in this case, .49 m. Then solve for Vf.Is there any way to close a thread?
 
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