Work Energy Theorem Question, Relatively Easy

[jcfor3ver]

Homework Statement

The left side of the figure shows a light (`massless') spring of length 0.320 m in its relaxed position. It is compressed to 74.0 percent of its relaxed length, and a mass M= 0.230 kg is placed on top and released from rest (shown on the right).

The mass then travels vertically and it takes 1.10 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.

Homework Equations

W= Ek
W= Eg
w = F d
Eg= mgh
Ek = 1/2mv^2

W=KEfinal-KEinitial
Elastic Spring Pot (Us)=-1/2k(displacement)^2
vf=vi+at

The Attempt at a Solution

first I found the displacement of the spring to be .2368, and that is my displacement squared for my Us.
Then I found the velocity (vi) to be 10.78 m/s (vi=9.8*1.10seconds)
Then i found the distance height (vi/2*t)= 5.929 m
Then I used the work energy theorem, I set the Us equation=1/2mv^2+mgh and solved for the spring constant k to get 39.39 N/m. But it is wrong, I have tried other ways but got unusually large numbers for my answer, which did not make sense. Help please? I have a test in a few hours.

 

[kuruman]

You don't need to go through the kinetic energy calculation. Just say that the initial elastic potential energy stored in the spring is equal to the final gravitational potential energy of the mass. That might make your calculation more transparent to trouble shoot.

 

[p21bass]

kuruman is correct. Also, your distance appears to be incorrect - what you used is the equation for average displacement.