# How much does the spring stretch?

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1. May 18, 2015

### kaspis245

1. The problem statement, all variables and given/known data
A mass $m$ body is attached to the spring and rests on the board so the spring is not stretched. The board starts to move with an acceleration $a$. How much does the spring stretch at the moment when the board separate from the body?

2. Relevant equations
Newton's Laws of Motion

3. The attempt at a solution
Here is my drawing:

I think that the board will separate from the body when $kx=m_1a$, where $m_1$ is the mass of the board. I can't use $m_1$ since it is not given in the problem but I can't think a way to replace it.

Is my reasoning correct and how can I replace $m_1$?

2. May 18, 2015

### BvU

Hello Kaspis. Funny you should draw a picture with three forces and then mention only two to find the separation. And these two work on two different objects!

Think about what the forces are on the mass m. There is a third one. And remember Newton 2.

3. May 19, 2015

### kaspis245

Here's the corrected image (please nevermind how the forces are represented, only directions matter here):

I know that $F_N=mg$, so $F_N$ should be 0 when $kx=mg$, therefore $x=\frac{mg}{k}$. Is that the answer?

Update:
Now that I think about it, the equation $x=\frac{mg}{k}$ should not suffice. The amount of stretch is clearly dependant on the acceleration with which the board is falling. For example, if the board is falling with many times bigger acceleration than $g$ it will surely separate from the board sooner than it would with smaller acceleration than $g$. Thus, there must be some kind of relationship between board's acceleration and other forces.

Last edited: May 19, 2015
4. May 19, 2015

### BvU

How do you know $F_N = mg$ ? That is only when the board is at rest at the point where the extension of the spring is zero. Once there is acceleration that can't be true any more.

5. May 19, 2015

### kaspis245

Sorry, I'll apply Newton's second law here:
$kx+F_N-mg=-ma_1$, where $a_1$ is the acceleration with which the body is moving. Unfortunately, this doesn't help at all .

Maybe I need to find the rate at which $F_N$ decreases and the rate at which $m_1a$ increases. Would that help?

6. May 19, 2015

### BvU

$a$ is a given acceleration. As soon as $F_N$ hits 0, the mass comes off the board. Only one unknown: $x$ !

7. May 19, 2015

### kaspis245

Oh, so the equation should be like this:
$kx+F_N-mg=-ma$

If $F_N$ is 0, then:
$kx-mg=-ma$
$x=\frac{m(g-a)}{k}$

Is that correct?

8. May 20, 2015

### BvU

What do you think ? Does it look plausible ?
- it goes to $kx = mg$ for $a$ goes to 0, so that seems OK
- if $a = g$ there is free fall from the start, so that looks OK too
- there are no strange outsiders contributing (such as $m_1$), which seems good

I'm all in favour of this outcome ! Well done.

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