# Homework Help: Spring, mass change.

1. Jan 26, 2009

### Jamboree

I feel like I must be missing something here. I think I'm following the right steps, yet my answer doesn't match one of the choices.

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

A Mass of 3.0kg is hung from a spring, causing it to stretch 12 cm at equilibrium. The 3.0kg mass is then replaced by a 4.0kg mass and the new block is released from the unstretched position. How far will the 4.0kg mass block fall before its direction is reversed?

a) 9 cm
b) 18 cm
c) 24 cm
d) 32 cm
e) 48 cm

2. Relevant equations
Spring Constant-
k= F/x

3. The attempt at a solution
First, I found the force of the 3kg block.
F= 3 * 9.8 = 29.4N
Next, I found the spring constant K
K= F/x[distance] = 29.4/.12m= 245

I Worked backwards to find the distance traveled by the 4kg block, like this.
K=F/x
245= (4*9.8)/x
x=.16m = 16 cm
Which is not one of the options. Where have I gone wrong?

Thanks for any help, it is much appreciated.

2. Jan 26, 2009

### LowlyPion

Welcome to PF.

When you replace the weight and release it, the potential energy of the weight, will become kinetic energy and will also start adding potential energy into the spring. The maximum distance it falls will be when all the gravitational energy becomes potential energy in the spring and kinetic energy becomes momentarily 0.

m*g*h = 1/2*k*h2

3. Jan 26, 2009

### nrqed

You are finding the new equilibrium position when the 4 kg is attached.
But they ask what happens when the 4kg is dropped. Then the srping will stretch way beyond the new equilibrium position. What you must use is conservation of energy in the form mgh = 1/2kx^2

4. Jan 26, 2009

### Jamboree

Wow, I'm amazed I managed to completely ignore that. It always seems so obvious in retrospect....I'm going to go ahead and blame the late hour and lack of sleep.

Thank you both for the help, I've got it now.