# Homework Help: Simple Spring problem

1. May 9, 2010

### IDumb

Hey guys, I have my physics final tomorrow and I have this simple question from a practice exam. I have gotten the correct answer by using the momentum principal, but when I try to check my work using the energy principal I get twice the correct answer. So I would appreciate it if you guys could help me figure out what I am mixing up in the energy principal. Thanks a lot.

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

A 5 kg block is attached to a spring with stifness = 1500 N/m and relaxed length 0.2 m. You hang the spring and mass from the ceiling. When the block is hanging motionless from the spring, what is the stretch?

3. The attempt at a solution

Using the momentum principal:
Pf-Pi=0=Fnet
Fnet=Fspring-Fgrav

Using the energy principal:
System: Block + Spring + Earth
Initial State: Block is released from rest, spring is relaxed.
Final State: Block is at rest, hanging from the spring.
*I assume that the final height of the block is 0, and the initial height is = to the stretch.* (I think this is where I am going wrong?)
Ef=Ei + Wsurroundings
Kf + Uf,spring + Uf,grav = Ki + Ui,spring + Ui,grav + 0
0 + (1/2)(Ks)(s^2) + mgh = 0 + 0 + mgh

*I assume that the final height of the block is 0, and the initial height is = to the stretch.* (I think this is where I am going wrong?)

(1/2)(1500)(s^2) = mg(s)
s=2mg/1500=.065333

Thanks again for the help

2. May 10, 2010

### collinsmark

Hello IDumb,

Welcome to Physics Forums!

Conservation of energy isn't the way you want to approach this problem. The disconnect here is that potential energy is not conserved in this problem.

I know it sounds a bit weird, but consider what would happen if you hold the weight in your hand while it is attached to the spring, such that the stretch is zero, and then you release the weight. The weight would fall down a little and the spring would stretch, until finally the velocity of the weight would reach zero. This is the situation where (1/2)ks2 = mgs. But it doesn't end there! The upward force on the spring is equal to 2mg (and only mg in the downward direction)! So the weight now accelerates upward, back to (almost) the point where the stretch equals 0, and the process repeats. You've created an example of simple harmonic motion.

But after awhile, due to air resistance, and other forms of friction, the system will eventually reach equilibrium at the midpoint (when the motion dies out). But by that time, much of the initial energy was lost to heat. And that's why you cant use conservation of potential energy on this particular problem. Half the energy goes into heat by the time the motion stops.

Last edited: May 10, 2010
3. May 10, 2010