AP Spring Problem

Right now I'm reviewing old AP problems for the AP exam for Physics B tomorrow, and I can't seem to get a simple spring problem. I don't know why, but I can't just grasp it. Here it is:

A block of mass 3kg is hung from a spring, causing it to stretch 12 cm at equilibrium, as shown above. The 3 kg block is then replaced by a 4 kg block, and the new block is released from the position shown above, at which the spring is unstretched. How far will the 4 kg block fall before its direction is reversed?

Here's my work(I'm allowed to use 10m/s^2 for gravity):
3kg(10m/s^2) + .12k = 0
.12k = 30
k = 250

250x - 4(10) = 0
x = .16 m = 16 cm

The answer I'm told is supposed to be 32... can someone explain what I'm doing wrong?
 

nrqed

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WrathofHan said:
Right now I'm reviewing old AP problems for the AP exam for Physics B tomorrow, and I can't seem to get a simple spring problem. I don't know why, but I can't just grasp it. Here it is:

A block of mass 3kg is hung from a spring, causing it to stretch 12 cm at equilibrium, as shown above. The 3 kg block is then replaced by a 4 kg block, and the new block is released from the position shown above, at which the spring is unstretched. How far will the 4 kg block fall before its direction is reversed?

Here's my work(I'm allowed to use 10m/s^2 for gravity):
3kg(10m/s^2) + .12k = 0
.12k = 30
k = 250

250x - 4(10) = 0
x = .16 m = 16 cm

The answer I'm told is supposed to be 32... can someone explain what I'm doing wrong?
What you found is the new *equilibrium position*. If the 4kg was slowly released so that it would not oscillate, this is where it will end up (try to visualize holding the mass and letting it go *very* slowly so that when you are not touching it anymore it remains at rest).

But in this question, the mass is dropped suddenly. So it oscillates. when it is released, it is 16 cm above its equilibrium position. This means that it will go all the way down to 16 cm *below* its equilibrium position before moving up again. So a total of 32 cm below its initial position
 
nrqed said:
What you found is the new *equilibrium position*. If the 4kg was slowly released so that it would not oscillate, this is where it will end up (try to visualize holding the mass and letting it go *very* slowly so that when you are not touching it anymore it remains at rest).

But in this question, the mass is dropped suddenly. So it oscillates. when it is released, it is 16 cm above its equilibrium position. This means that it will go all the way down to 16 cm *below* its equilibrium position before moving up again. So a total of 32 cm below its initial position
In Physics B, would the max distance always be twice the equilibrium point?
 

Andrew Mason

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WrathofHan said:
Right now I'm reviewing old AP problems for the AP exam for Physics B tomorrow, and I can't seem to get a simple spring problem. I don't know why, but I can't just grasp it. Here it is:

A block of mass 3kg is hung from a spring, causing it to stretch 12 cm at equilibrium, as shown above. The 3 kg block is then replaced by a 4 kg block, and the new block is released from the position shown above, at which the spring is unstretched. How far will the 4 kg block fall before its direction is reversed?

Here's my work(I'm allowed to use 10m/s^2 for gravity):
3kg(10m/s^2) + .12k = 0
.12k = 30
k = 250

250x - 4(10) = 0
x = .16 m = 16 cm

The answer I'm told is supposed to be 32... can someone explain what I'm doing wrong?
The potential energy of the 4 kg block is converted into potential energy of the spring.

[tex]mgx = \frac{1}{2}kx^2[/tex]

So:

[tex]x = 2mg/k[/tex]

[tex]x= 2*4*10/250 = .32[/tex]

AM
 

nrqed

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WrathofHan said:
In Physics B, would the max distance always be twice the equilibrium point?
If the object is dropped (it has no initial velocity) then the distance from the highest point to the lowest point is twice the ditance between the highest point and the equilibrium position, yes (the mass follows simple harmonic motion, that is a cosine type of curve)

You can also solve the problem using conservation of energy as Andrew Mason posted (the equation he gave works because the initial and the final velocities are both zero in that example)
 
Cool, thanks, that helped a lot =)
 

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