# Spring Maximum Length

1. Mar 26, 2015

### Westin

1. The problem statement, all variables and given/known data

A mass of 0.36kg hangs motionless from a vertical spring whose length is 0.8m and whose unstretched length is 0.51m. Next, the mass is pulled down to where the spring has a length of 1.09m and given an initial speed upwards of 1.2m[PLAIN]https://s3.lite.msu.edu/adm/jsMath/fonts/cmmi10/alpha/144/char3D.pngs. [Broken] What is the maximum length of the spring during the motion that follows?

Maximum length=

2. Relevant equations

F = k*y
1/2*k*A^2

3. The attempt at a solution

k = F/y = m*g/y = 0.36kg*9.8m/s^2/(0.8m - 0.51) = 12.17793N/m

Energy is conserved

Let A be the maximum displacement

So 1/2*k*A^2 = 1/2*k*(1.09 - 0.51)^2 + 1/2*m*v^2

so A = sqrt((1.09 - 0.51)^2 + m/k*v^2) = sqrt((1.09 -0.51)^2 + 0.36/12.18*1.2^2) = 0.5974m

So the length will be 0.51 + 0.5974m = 1.11m

I only have one attempt wrong, Cant find the error..

Last edited by a moderator: May 7, 2017
2. Mar 26, 2015

### BvU

You only have two relevant equations -- if I count the second as an equation: spring potential energy = 1/2 k x2.

You also use a third one: kinetic energy = 1/2 m v2.

You forget a fourth one. Yet another energy....

Oh, and: perhaps you want to check "1/2*k*(1.09 - 0.51)^2" which doesn't look right to me. At what point do you expect there is no energy stored in the spring ? And is that the top position, or is that the middle between top and bottom of the oscillation ?

Last edited: Mar 26, 2015
3. Mar 26, 2015

### Westin

Would it be 1/2*k*(1.09 - 0.8)^2 and 1/2*k*(.8 - 0.51)^2

I do you KE=1/2 mv^2 I thought? I'm confused on what else there could be.

4. Mar 27, 2015

### BvU

Check your energy approach with the following:
Hold the weight at spring unextended length 0.51 m and let go. How far will it drop until it will go up again (i.e. v = 0)
Would you do ${1\over 2} kx^2 = {1\over 2} mv^2= 0$ to find x ?