Spring Constant Calculation for Compressed Spring with Mass

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michlip
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1. Homework Statement
The left side of the figure shows a light (`massless') spring of length 0.340 m in its relaxed position. It is compressed to 67.0 percent of its relaxed length, and a mass M= 0.250 kg is placed on top and released from rest (shown on the right).

The mass then travels vertically and it takes 1.50 s for the mass to reach the top of its trajectory. Calculate the spring constant, in N/m. (Use g=9.81 m/s2). Assume that the time required for the spring to reach its full extension is negligible.


2. Homework Equations
EK=1/2mv2=1/2 * 0.25 * V2
K=-0.5mv2/x

3. The Attempt at a Solution
-0.5mv2/x
v=14.715
x=11.22
m=0.250
I plugged in the variables, but it still isn't right.
 
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The mechanical energy of the spring will be converted into gravitational potential energy at the top of the trajectory. You have used kinetic energy instead.
 
im really confused by this same type of question...
i found U(y) now how do i use that to find K?
 
As someone above said, you should be converting mechanical energy into gravitational potential energy, rather than using kinetic. 1/2kx^2=mgy is the proper conversion of energy. Hope that helps
 
is x = .2278 m like above? and y = the distance the mass traveled in the air? because i have been using those and gettting the wrong answer
 
^

The x i used in my equation is the distance that the spring moves from its "position of relaxation". You need to find the distance that the object goes vertically, and plug that in for y in the equation

1/2kx^2=mgy >>>>> k= (mgy)/(.5x^2)

Once you have y you can solve
 
If you can confirm that the asnwer is roughly 3.1E3 N/m , I will tell you how i got it.

I believe in the equation x= x(initial) + Velocity(Initial)*time + .5gt^2 you forgot to account for the .5gt^2, which equals roughly 22 m. Added to the 11 m you found using the middle term of the equation above, you have solved for the correct height. You were only a step off from getting it.