Spring-block problem

1 vote
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I am told to find acceleration of masses $m_2$ and $m_1$ after the wall/support is removed. My teacher has told me that the spring constant won't change instantaneously and we can treat it as constant.

My attempt

Due to the force $F$ the block will press the spring till equilibrium is achieved which gives me:

$F=kx$ ( k is spring constant and x is compression)

$x=F/k$

Now when the support is released then that's where I am getting confused over some parts.

Will the removing of support produce an acceleration on $m_2$ just because the compression of the spring by $m_1$ will result in the disturbance from equilibrium point and the spring will try to achieve it's original length by expanding the same length ? This will make sense but can't the spring push back $m_1$ again?

How should I analyse this motion?

edited Jul 18, 2018

1 vote

The following diagrams show the forces (red) exerted on the blocks (blue). I have not included the forces which the blocks exert, because these are not relevant for finding the accelerations of the blocks.

I. Before the wall is removed

Let the compressive force in the spring be $T$. Both masses are in static equilibrium. The forces $F$ and $T$ are the only horizontal forces acting on $m_1$ (I assume there is no static friction) and they act in opposite directions. Therefore they are equal : $T=F$.

Likewise $m_2$ is also in static equilibrium, acted on by the spring force $T$ to the right and the reaction force $N$ from the wall to the left. Therefore $N=T=F$.

II. Immediately after the wall is removed

Your teacher is correct : the compressive force $T$ in the spring does not change instantaneously when the wall on the right is removed. The length of the spring has not yet changed at this instant in time, so the compressive force $T$ has the same value it had before.

The same 2 forces still act on $m_1$, so the resultant force on it is still zero, hence the acceleration of $m_1$ is zero.

The wall is no longer providing the reaction force $N$ which kept $m_2$ in static equilibrium. So the resultant force on $m_2$ is now the compressive force $T=F$ acting to the right. The acceleration of $m_2$ is $F/m_2$.

III. A short time after the wall is removed

After $m_2$ has accelerated and moved a short distance, the spring is now longer so the compression force in it is smaller : if has reduced from $F$ to $F'$.

The forces on $m_1$ are no longer balanced because the spring force $F'$ is now smaller than the applied force $F$, so there is a resultant force on $m_1$ hence$m_1$ now accelerates to the right at a rate of $(F-F')/m_1$.

The resultant force on $m_2$ is now $F'<F$ so the acceleration of block $m_2$ is now $F'/m_2$.

answered Jul 18, 2018 by (28,466 points)
edited Jul 25, 2018
So the acceleration of m1 is zero because the spring doesn't change and m2 feels a spring force because of it's compressed state ,right
That is correct.
But why ism2 experiencing spring force when it is not pushing the spring .........is it application of newtwon's third law....this is my only confusion
$m_2$ is experiencing spring force and is pushing back against the spring, both before the wall is removed and afterwards. Yes this is an application of Newton's 3rd Law.