Solve Spring Mass Change Problem: Find Distance for 4kg Mass

[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.

Homework Statement

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

Homework Equations

Spring Constant-
k= F/x

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.

 

[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

 

[nrqed]

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.

Homework Statement

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

Homework Equations

Spring Constant-
k= F/x

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.

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

 

[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.