Why can we move the spring with constant speed when we apply a force?

[Lotto]

Homework Statement

A block of mass $m$ is attached to a spring of an initial lenght $l_0$ with a spring constant $k$. The other end of the spring is moved with a constant velocity $v$ away from the block. Find the maximum lenght of the spring.

Relevant Equations

$\frac 12 k{\Delta l}^2=\frac 12 m v^2$

One solution is that if we move with the same velocity as the spring does, the initial velocity of the block will be $v$ and the final will be zero, so its kinetic energy will transform into a potential energy of the spring.

I would also say that we can say that if we pull the spring, we apply a force $F=k\Delta x$ on the spring as well as on the block, so in all, we do a work $\frac 12 k{\Delta l}^2$ that is equal to the spring's potencial energy and to the block's kinetic energy.

But my problem is: how is it possible to move the spring with a constant velocity when we apply a force? It should accelerate then. My idea is that it is because the spring is probably massless, but I am not sure.

What is the reason?

If the block had a mass $M$ and the spring had a mass $m$, what would be the equation? Then we couldn't move the spring with a constant speed?

 

[Aurelius120]

You can't move an object at rest without a force but you can make net force zero once it moves with desired velocity.
Apply a force larger than the opposing spring force. The system gains an acceleration and reaches a velocity where the net acceleration is zero because opposing spring force is equal to supplied force but there is velocity.

 

[Orodruin]

The force on the other end of the spring will not be constant.

 

[PeroK]

If the block had a mass $M$ and the spring had a mass $m$, what would be the equation? Then we couldn't move the spring with a constant speed?

If you had to set up this scenario as an experiment to test the theory, how would you do it?

 

[Chestermiller]

The spring is not moving with constant velocity v. The far end of the spring is moving with constant velocity v. Check the problem statement.

 

[Lotto]

So if I understand it well, if I have the spring in my hand and act on it with a force $F=k\Delta x$, the spring acts on my hand with an opposite force, so the net force acting on my hand is zero, therefore my hand and the furthest point of the spring moves with a constant speed.

And the speed is not the same for the whole spring, every point has a different speed. It is similar to the ant on a rope paradox.

 

[Orodruin]

So if I understand it well, if I have the spring in my hand and act on it with a force $F=k\Delta x$, the spring acts on my hand with an opposite force, so the net force acting on my hand is zero, therefore my hand and the furthest point of the spring moves with a constant speed.

And the speed is not the same for the whole spring, every point has a different speed. It is similar to the ant on a rope paradox.

This is a very common misunderstanding of Newton’s third law. The force from the hand on the spring does not act on the hand - it acts on the spring - and therefore is not part of the force balance for the hand.

 

[Lotto]

This is a very common misunderstanding of Newton’s third law. The force from the hand on the spring does not act on the hand - it acts on the spring - and therefore is not part of the force balance for the hand.

I meant that my body generates a force on the hand , and the hand acts with the force on the spring. The spring acts on the hand with the opposite force, so the net force acting on the hand is constant.

 

[Orodruin]

If the hand moves at constant speed, then the net force on the hand is zero. Whatever force the spring acts on the hand with, it must be countered by a force at the wrist. However, neother of those forces need to be constant.

 

[Steve4Physics]

A few thoughts...

One solution is that if we move with the same velocity as the spring does, the initial velocity of the block will be $v$ and the final will be zero,

That's a nice approach - you observe the system from an inertial frame of reference moving with velocity $\vec v$; in this frame of reference, one end of the spring is always stationary. But note that the initial velocity of the block in this frame is reference is $-\vec v$ (not $\vec v$).

Have you met simple harmonic motion (SHM)? That's what you will see in the moving frame of reference. The energy cyclically swaps between kinetic and potential.

Do you think doing th experiment with the apparatus in a moving truck (of large mass) would work (or, for vertical motion, in an elevator (of large mass))?

A far as dealing with a spring of significant mass, you will need to learn about SHM and use a bit of calculus (as different bits of the spring are moving with different velocities). But it turns out quite simply. Look up 'effective mass of a spring' if interested.

 

[Aurelius120]

The spring is not moving with constant velocity v. The far end of the spring is moving with constant velocity v. Check the problem statement.

By far end, you mean the end away from the hand or away from the block? Also if only one end moves with constant velocity, does it mean the spring is getting compressed?

 

[PeroK]

By far end, you mean the end away from the hand or away from the block?

It says the end away from the block in the problem statement.

Also if only one end moves with constant velocity, does it mean the spring is getting compressed?

Extended, not compressed. Again, see the problem statement.

 

[kuruman]

We are asked to "find the maximum length of the spring." I interpret this to mean that, as the free end of the spring is pulled by, say, a disembodied hand at constant velocity, the mass oscillates and the spring has a maximum and a minimum length. We are asked to find the former.

 

[Orodruin]

Extended, not compressed. Again, see the problem statement.

Both, actually. In the rest frame of the end moving at constant velocity, the mass will perform SHM meaning the spring will alternate being extended and compressed. Actually, for the problem question, it doesn’t matter if the end starts moving toward or away from the mass.

 

[Lotto]

So the speed at the end is $v$ because when I act with a force on the spring, stretching it, the spring "wants" to get into its initial position, so it acts with an oppsite force equal to $F=k\Delta x$. Is it correct?

But I am not sure why the body on the spring should oscillate? I imagined it the way that when the body gains the speed $v$, it moves with this speed without any oscillation. Why is it wrong?

 

[PeroK]

So the speed at the end is $v$ because when I act with a force on the spring, stretching it, the spring "wants" to get into its initial position, so it acts with an oppsite force equal to $F=k\Delta x$. Is it correct?

But I am not sure why the body on the spring should oscillate? I imagined it the way that when the body gains the speed $v$, it moves with this speed without any oscillation. Why is it wrong?

I think you need to get into a lab and start experimenting. Why does a ball bounce off the floor? Why does a pendulum swing back and forwards? Why does a mass on a spring oscillate?

Okay, you can predict this stuff from Newton's laws. But it's sure a helluva lot easier if you have some physical intuition about these scenarios.

 

[Lotto]

And will the body start to oscillate after I streched the spring on a maximum lenght? Because it seems odd to me for the body to oscillate during the stretching.

 

[PeroK]

And will the body start to oscillate after I streched the spring on a maximum lenght? Because it seems odd to me for the body to oscillate during the stretching.

I don't know what that means. If the spring is stretching, then the block cannot be oscillating. An oscillating block, in this problem, means an alternating sequence of stretching and compressing of the spring. With, if you do the maths, simple harmonic motion.

 

[kuruman]

Here's one way to look at it.
Suppose you see the spring-mass system moving on a frictionless platform moving at constant velocity $v$ so that the spring is stretched by a constant amount $\Delta x.$ I enclose the platform with walls and a ceiling and put you inside it. You cannot look outside and you cannot tell that the platform is moving at constant velocity because you are at rest with respect to it and there is no net force acting on either you or the mass. Yet, you see that the spring is stretched and stays that way. Wouldn't that seem odd to you?

Here is another way to look at it.
We are told that the mass starts from rest and then starts moving at constant some velocity $\cancel{v}$ that at some point in time matches the constant velocity of the free end of the spring. For the velocity of the mass to change, there must be an acceleration. For the acceleration to be non-zero there must be an external force $F$ at the free end of the spring. When $F$ is "turned on", the spring starts stretching and there are two external forces acting on the mass spring, $F$ and the spring force from the mass. At that point the mass will start oscillating. If you don't believe me, do the experiment as others have suggested.

• Hang a mass vertically at the end of a spring and hold it at the unstretched position of the spring.
• Whenever the spirit moves you, turn on constant force $F=mg$ by letting the mass drop.
• Watch what happens.

Note: This post has been edited to fix the paragraph starting with "We are told $\dots$" See posts #20 and #21.

 

[nasu]

We are told that the mass starts from rest and then is moving (eventually) at constant velocity $v$.

Are we told this? When the block reaches the velocity $v$ the spring has the maximum extension.

 

[kuruman]

Are we told this? When the block reaches the velocity $v$ the spring has the maximum extension.

I used "constant velocity $v$" as an identifier for the velocity of the free end of the spring. I admit it was bad English, so I edited my post to clarify.

 

[nasu]

Also, you seem to say that there are two forces acting on the block, the spring elastic force and another one you call "F". Is this just me, missunderstanding your explanation? If not, where does this come from? You say that there is another force, also called "F" acting at the other end of the spring, the one moving with constant velocity.

 

[kuruman]

Also, you seem to say that there are two forces acting on the block, the spring elastic force and another one you call "F". Is this just me, missunderstanding your explanation? If not, where does this come from? You say that there is another force, also called "F" acting at the other end of the spring, the one moving with constant velocity.

Consider the FBD of a point P on the end of the spring that is moving at constant velocity $v$. The point has zero acceleration. If the spring is extended, we know that at P there is a spring force of magnitude $k|x|$ directed opposite to the velocity tending to contract the spring. It follows that there must be an additional force $F$ of equal magnitude opposing the spring force in order to make the acceleration of P equal to zero.

To put it simply, you cannot stretch a spring unless you pull on both ends. So here, we have to assume the existence of some entity, say a disembodied hand, moving at constant velocity $v$ while the mass-spring system executes its oscillations. If we assume no entity at the other end, the only way I see for point P to have zero acceleration is having the spring moving as one with the mass at constant $v$ with its ends separated by the equilibrium length $l_0$.

 

[Lnewqban]

We are asked to "find the maximum length of the spring." I interpret this to mean that, as the free end of the spring is pulled by, say, a disembodied hand at constant velocity, the mass oscillates and the spring has a maximum and a minimum length. We are asked to find the former.

The shown relevant equation seems to imply that not friction exists.
If that is the case, could you please explain why is an oscillation of the mass induced?
Would it still happen for very low values of hand constant velocity?

 

[nasu]

Consider the FBD of a point P on the end of the spring that is moving at constant velocity $v$. The point has zero acceleration. If the spring is extended, we know that at P there is a spring force of magnitude $k|x|$ directed opposite to the velocity tending to contract the spring. It follows that there must be an additional force $F$ of equal magnitude opposing the spring force in order to make the acceleration of P equal to zero.

To put it simply, you cannot stretch a spring unless you pull on both ends. So here, we have to assume the existence of some entity, say a disembodied hand, moving at constant velocity $v$ while the mass-spring system executes its oscillations. If we assume no entity at the other end, the only way I see for point P to have zero acceleration is having the spring moving as one with the mass at constant $v$ with its ends separated by the equilibrium length $l_0$.

I agree that there are two forces acting on the end (point P) moving with constant velocity. But you said that there are two forces acting "on the mass". You also say "that the mass starts from rest" so it seems by "the mass" you mean the mass at the other end, not at point P. There is only one force acting on it, isn't it?

 

[Orodruin]

The shown relevant equation seems to imply that not friction exists.
If that is the case, could you please explain why is an oscillation of the mass induced?

Go to the rest frame of the end of the spring moving at constant velocity. In that frame the mass starts with velocity -v connected to a spring initially unstretchef, which results in shm.

Would it still happen for very low values of hand constant velocity?

Yes. What would change is the amplitude of oscillation.

 

[kuruman]

I agree that there are two forces acting on the end (point P) moving with constant velocity. But you said that there are two forces acting "on the mass".

I see that I did and that was a mistake that I overlooked when I edited the post. It is fixed now. Thanks.

You also say "that the mass starts from rest" so it seems by "the mass" you mean the mass at the other end, not at point P. There is only one force acting on it, isn't it?

Yes, I mean the mass of the block and yes there is only one force acting on it if "it" refers to the block.

 

[kuruman]

The shown relevant equation seems to imply that not friction exists.

I agree and I assume no friction between the block and the surface on which it slides.

If that is the case, could you please explain why is an oscillation of the mass induced?

Gladly.

• We are told that the leading end of the spring moves at constant velocity.
• The spring is the only entity that can exert a horizontal force on the block, therefore any force it exerts on the block is actually the net force on the block that causes it to accelerate.
• If the spring is at maximum extension $x_{\text{max}}$, the block is at a turning point. Maximum extension means that the two ends of the spring are instantaneously at rest with respect to other. It follows that they each have velocity $v$ relative to the surface. Also, at this turning point the acceleration of the mass has the maximum value $a_{\text{max}}=\dfrac{kx_{\text{max}}}{m}$. It follows that past the turning point the velocity of the block and the trailing edge of the spring will increase from $v$ and they will start approaching the leading edge of the spring.
• When the distance between the two ends of the spring is $l_0$, the block will instantaneously stop accelerating, but it will be moving faster than the leading end.
• Past that point the block and trailing end will slow down relative to the leading end until their velocity, once again, becomes $v$ at the maximum compression length.

I believe I just described half an oscillation cycle.

 

[Lnewqban]

Many thanks to both, @Orodruin and @kuruman

 

[kuruman]

For future reference and to tidy this up a little bit, this problem has a straightforward solution in the lab frame. Here is an outline.

• Choose an arbitrary origin and define coordinates $x_l$ and $x_t$ for the leading and trailing ends of the spring respectively. Note that $x_t$ is also the position of the block.
• Write an equation for $x_l$ as a function of time (trivial).
• Find an expression for change in the spring's length from the equilibrium position and hence the force exerted by the spring on the block.
• Solve the differential equation obtained from Newton's second law to find $x_t$ as a function of time.
• Maximize $x_l-x_t.$

I found the answer a bit surprising.