Mass on a vertical string - Conservation of Energy Problem

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Bryon
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Homework Statement



https://wug-s.physics.uiuc.edu/cgi/courses/shell/common/showme.pl?cc/DuPage/phys2111/fall/homework/Ch-08-GPE-ME/mass_vertical_spring/7.gif

A spring with spring constant k = 45 N/m and unstretched length of L0 is attached to the ceiling. A block of mass m = 1.5 kg is hung gently on the end of the spring.

a) How far does the spring stretch? Answer = 0.327m

Now the block is pulled down until the total amount the spring is stretched is twice the amount found in part (a). The block is then pushed upward with an initial speed vi = 2 m/s.

b) What is the maximum speed of the block? <---this one I am not sure of.

Homework Equations



0.5*m*v^2 = Ke
0.5*k*x^2 = K(spring)
U + K = Uo + Ko

The Attempt at a Solution



0.5*m*v(final)^2 = 0.5*m*v(initial)^2 + 0.5*k*x^2

for v I got 4.1 m/s. But the answer is incorrect. Any ideas where I went wrong?
 
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Your method for (b) is correct. At first I did what you did and got the same numbers as you. Then I reread the problem carefully. It is not very clear, but when it says "the block is pulled down until the total amount the spring is stretched is twice the amount found in part (a)", it means the spring is stretched from the unstretched position by twice the amount not from the new equilibrium by twice the amount. Try it.
 
Ah, ok so what I want to find is the length (L) of the relaxed spring with no force applied to it?
 
Bryon said:
Ah, ok so what I want to find is the length (L) of the relaxed spring with no force applied to it?
You don't need to find that, in fact you can't find that. All I'm saying is that the x in your last equation should be 0.327 m not 2*0.327 m. It represents the displacement from equilibrium when the mass is hanging and that is 0.327 m.
 
Ah, ok I had to read that a few times to understand it. That question is rather confusing, but I think i now see why its not twice the distance. Again, thanks for the help I truly appreciate the guidance!