Calculating Coefficient of Static Friction to Keep Glider from Springing Back Left

In summary, an air-track glider with a mass of 0.100 kg and a spring with a force constant of 20.0 N m is initially moving at 1.50 m/s to the right on a horizontal air track. When the air track is turned off, the glider travels 8.6 cm before stopping. To prevent the glider from springing back to the left, the coefficient of static friction would need to be greater than a certain value. If the coefficient of static friction is 0.60, the maximum initial speed the glider can have while remaining at rest after stopping is being solved for, using the equations Wtotal = -(1/2)kd^2 - mgd*u and
  • #1
henry3369
194
0
An air-track glider of mass 0.100 kg is attached to the end of a horizontal air track by a spring with force constant 20.0 N m (Fig. 6.22a). Initially the spring is unstretched and the glider is moving at 1.50 m s to the right.
It is calculated that with an air air track turned off, a glider travels 8.6 cm before it stops instantaneously. How large would the coefficient of static friction have to be to keep the glider from springing back to the left? (b) If the coefficient of static friction between the glider and the track is 0.60 what is the maximum initial speed that the glider can be given and still remain at rest after it stops instantaneously? With the air track turned off, the coefficient of kinetic friction is 0.47.

I need help with part b.
I know how to solve this but I'm getting the wrong answer because of the sign of the work done.

In my book is shows that to solve part b you use:
total work = Wspring + W fric = change in KE.

What I don't understand is why my book made both the work of the spring and work of friction negative if their forces are acting in opposite directions. After the glider stops, the block will move the left and the force of the spring will be to the left because it is restoring, and the force of friction would be to the right because the friction opposes the motion, so wouldn't the work done by the friction be negative because it points in the opposite direction of the displacement?
 
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  • #2
Hi. This post should be in Homework and Coursework questions, I think..
But I think that what they are getting at is to do with how the force on the spring relates to the energy stored. What happens after the slider stops initially will depend upon whether or not the spring has been stretched appropriately for it to pull the slider back (overcoming the friction) or just stuck to the track and the force from the spring being too little. Work does not have a direction (scalar) so it is not 'wrong' to have two different signs for the forces if work is being done on Friction and Stretching the Spring.
 
  • #3
sophiecentaur said:
Hi. This post should be in Homework and Coursework questions, I think..
But I think that what they are getting at is to do with how the force on the spring relates to the energy stored. What happens after the slider stops initially will depend upon whether or not the spring has been stretched appropriately for it to pull the slider back (overcoming the friction) or just stuck to the track and the force from the spring being too little. Work does not have a direction (scalar) so it is not 'wrong' to have two different signs for the forces if work is being done on Friction and Stretching the Spring.
Sorry. I was not sure if this belonged in the homework section because I already knew the solution, but I had a hard time understanding what is going on. You said that it is not wrong to have two different signs which I understand, but I don't understand why it does not have two different signs in the problem. Shouldn't Work done by friction and work done by the spring have different signs? My book states that Wtotal = -(1/2)kd^2 - mgd*u(coefficient of kinetic friction) and I don't know why it is not Wtotal = (1/2)kd^Y2 - mgd*u.
 
  • #4
If it helps, my book also provides a hint: "Apply F = ma to fidn the maximum amount the spring can be compressed and sdtill have the spring force balanced by friction. Then use Wtotal = delta(KE) to find the initial speed that results in this compression of the spring when the glider stops."
 
  • #5
My understanding is that the distance of compression is used to find the total work with the force of the spring acting as a restoring force to the left while the force of kinetic friction to the right because it opposes the motion of the particle as it travels that distance to the left.
 
  • #6
henry3369 said:
My understanding is that the distance of compression is used to find the total work with the force of the spring acting as a restoring force to the left while the force of kinetic friction to the right because it opposes the motion of the particle as it travels that distance to the left.

I am trying to sort this out properly this time:
Afaics, those two forces are in the same direction (they both slow the slider down). I think that you either have misread the wording in your book or it is badly written (quite possible!). You refer to the work "of" the spring. Perhaps you would find it easier to consider work done by the slider in both cases. Forces against the friction and spring.
Work done against friction is Force (to the right) times distance moved (to the right) = positive work.
Work done against spring is Stretching Force (to the right) times distance moved (to the right) = positive work (=kd2/2)

These are equated to KE for one equation.
Then you have another equation for the static friction and the restoring force. That's the limiting case - which is what you are after.
Those two equations should be enough to give you the KE and hence the initial speed.
 
  • #7
henry3369 said:
An air-track glider of mass 0.100 kg is attached to the end of a horizontal air track by a spring with force constant 20.0 N m (Fig. 6.22a). Initially the spring is unstretched and the glider is moving at 1.50 m s to the right.
It is calculated that with an air air track turned off, a glider travels 8.6 cm before it stops instantaneously. How large would the coefficient of static friction have to be to keep the glider from springing back to the left? (b) If the coefficient of static friction between the glider and the track is 0.60 what is the maximum initial speed that the glider can be given and still remain at rest after it stops instantaneously? With the air track turned off, the coefficient of kinetic friction is 0.47.

I need help with part b.
I know how to solve this but I'm getting the wrong answer because of the sign of the work done.

In my book is shows that to solve part b you use:
total work = Wspring + W fric = change in KE.

What I don't understand is why my book made both the work of the spring and work of friction negative if their forces are acting in opposite directions. After the glider stops, the block will move the left and the force of the spring will be to the left because it is restoring, and the force of friction would be to the right because the friction opposes the motion, so wouldn't the work done by the friction be negative because it points in the opposite direction of the displacement?
I have this gut feeling that the book you are referring to, has misconceptions. I might be wrong, going to have to see the book.

For one thing: the sign of the work depends on your choice of system; that is: YOU decide relative to whom do you account the work done.

Think of work as gain or loss of energy through the action of a force.

You must choose the body or bodies you observe as your system.

For example:
If you push a box at a constant speed, frictional force converts all of the kinetic energy YOU give to the box into heat VERY FAST, so fast that in all practical purposes the box is not accelerating. Heat is really nothing more than trembling of the atoms in DIFFERENT random directions. Had there been no friction, all of the atoms would accelerate in one direction with equal accelerations, but since there is an interaction between the floor and the box (the frictional force), all of the energy you give to the box, instead of accelerating all of the atoms in one direction with equal accelerations, it will accelerate all of the atoms in different chaotic directions.
That is to say: friction makes the GAIN in kinetic energy do something else than accelerating all of the atoms in one single direction.

In other words: the frictional force converts the so-called mechanical energy into heat (by bending the irregularities of the bottom of the box).
BUT: that is not work. Converting one energy into another is NOT WORK. Because no energy gain or loss has occurred: the friction has not taken anything away or given anything.

BUT at the same time, friction is also stretching the floor; with this work IS DONE, because the energy is coming from the box - it LOSES energy and gives it to the floor due to friction.

Now, the SIGN of this work depends on your choice of system.

If you chose the box as the system, then the friction does NEGATIVE work, because it has taken energy away from the box.
If you chose the FLOOR as the system, then the friction has done POSITIVE work, because the floor GAINS energy due to the friction.

So conceptional reasoning might help solving this apparent paradox.

Can you give me the name of the book and the page this problem is in?
 
  • #8
It strikes me that the best way to consider this it that the KE of the moving slider is the Source of energy, so the slider does the work to suggest that Friction does work is like saying a resistor produces negative energy into a circuit; accurate as far as the sign goes but not very satisfactory as a concept.
 
  • #9
sophiecentaur said:
It strikes me that the best way to consider this it that the KE of the moving slider is the Source of energy, so the slider does the work to suggest that Friction does work is like saying a resistor produces negative energy into a circuit; accurate as far as the sign goes but not very satisfactory as a concept.

I am not sure that electric resistance is a good analogy to friction! Electric resistance is more like simply ALL OF THE circumstances together that make a current smaller in comparison to the situation where there were no such circumstances.
For example: if the wire is twice as long, the electric field is twice as weak, so the current is twice as weak.
If the wire has poor conductivity (eg few free electrons per unit volume), then the current, again, is smaller, then it would have been had the wire been made of different kind of lattice.
If the wire is made thinner, then there are less charge carriers per cross sectional area, and again, this makes the current smaller everywhere in the circuit.
If the temperature of the metallic lattice is higher, then the extra collisions with the electrons make probability of the electrons create the current smaller, this is again a factor that makes the current less..

Resistor is simply a part of the circuit that contributes the most for those factors. Note that half of the factors have nothing to do with... friction.. more like electric fields and geometry...

As for the minus sign: minus sign does not imply negative energy in the case of work, it rather implies that the system is LOSING energy due to some force doing work.
Friction CAN most certainly do work: positive AND negative, depending on the situation and YOUR CHOICE of system.
 
  • #10
Ott Rovgeisha said:
Electric resistance is more like simply ALL OF THE circumstances together that make a current smaller
Mmm. Not sure about that. When you run an electric motor, there may be 1kW of work done by that motor yet the resistive losses in the motor may be only be 100W. As far as the Power source is concerned, it 'sees' a 1.1kW load. If the motor and gearbox are only 80% efficient then there will be 200W of friction losses and 800W worth of useful work done. Resistance and Friction are both causes of Energy Loss to the system. The Resistance is certainly not the only 'circumstance' to limit the current flowing into the motor. Stall the motor and you will get a lot more current - and maybe burn the motor out.
 
  • #11
sophiecentaur said:
Mmm. Not sure about that. When you run an electric motor, there may be 1kW of work done by that motor yet the resistive losses in the motor may be only be 100W. As far as the Power source is concerned, it 'sees' a 1.1kW load. If the motor and gearbox are only 80% efficient then there will be 200W of friction losses and 800W worth of useful work done. Resistance and Friction are both causes of Energy Loss to the system. The Resistance is certainly not the only 'circumstance' to limit the current flowing into the motor. Stall the motor and you will get a lot more current - and maybe burn the motor out.

I never said that the resistance is the only cirmumstance that limits the current.
Similarly, the fact that a motor is overheating, doesn't necessarily mean that there is more current going...

If you stall the motor, then it essentially becomes a bundle of wires, the magnetic force is equalized with the force of whatever it is that is stalling the motor; so all of the energy the current carries ends up making atoms vibrate in a random motion: that means the energy takes the form of heat instead of rotation...That on itself causes it to heat up.

There may be some induction related things, but they rarely make the current bigger in a coil..
So when they are gone, that may make the current bigger. I am not sure here.. But even inductive effects are considered part of electrical resistance.
I may be wrong, but you seem to decide what is negative work or positive work by judging the TYPE of energy: whether it is the rotational energy of the motor or heat...
You should not assess the work done by energy conversion from one type to another, rather you should assess it by accounting for the energy gains or losses...
If the energy of the system is converted to heat energy, that does not mean negative work! Because that on itself does not mean that the wires LOST energy.. They did not lose it, it just got converted from one type into another. You must carefully watch if any gains or losses are there in QUANTITY, not in what forms energy takes.

A motor running of course encounters friction, but this is different from electrical resistance, fundamentally very very different...
Electrical resistance takes sort of an abstract form even: there is nothing there that actively RESISTs the current, or hinders it..
In connection to that: one must be vary careful while saying things like: the more resistance the current has the more the wire heats up, because this is not true.
If you connect a copper wire (a long one) to the battery, you will see that it heats up WAY more quickly and therefore stays WAY hotter than for example a nichrome wire.
I am assuming that the copper wire is long enough as to not short circuit the system.
 
  • #12
Ott Rovgeisha said:
is overheating, doesn't necessarily mean that there is more current going
Apart form the lack of cooling from its fan, what else can be causing it?
Ott Rovgeisha said:
If you stall the motor, then it essentially becomes a bundle of wires, the magnetic force is equalized with the force of whatever it is that is stalling the motor; so all of the energy the current carries ends up making atoms vibrate in a random motion: that means the energy takes the form of heat instead of rotation...That on itself causes it to heat up.
This is just plain wrong. You should look at some theory about the way motors work and the part that 'back emf' has to play in the process of limiting current through a motor. (Google can help you there and in other respects.)
 
  • #13
henry3369 said:
what is the maximum initial speed that the glider can be given and still remain at rest after it stops

henry3369 said:
After the glider stops, the block will move the left
No, after the glider stops, it stays stopped. Yes, the spring and friction will act in opposite directions then, but no further work is done.
henry3369 said:
total work = Wspring + W fric = change in KE.
The work discussed there is while the glider is still moving to the right. In that stage, friction and spring both act to the left.
 
  • #14
sophiecentaur said:
Apart form the lack of cooling from its fan, what else can be causing it?

This is just plain wrong. You should look at some theory about the way motors work and the part that 'back emf' has to play in the process of limiting current through a motor. (Google can help you there and in other respects.)

I have a feeling that you do not even read through what I posted. Selectively caught a sentence...

Even this sentence is NOT wrong at all.

If you stall the motor then two thing will happen:

1. The coils essentially DO become ordinary wires and instead of the rotational kinetic energy it would have gained if the magnetic force acted upon it like it did before the stalling, it now gains heat! That is simply the fact that the energy gain takes a different form!
2. If the motor stops rotating, that means that the coils do not experience a change in the magnetic field anymore (because the angle they are inside the magnetic filed caused by the magnets the coil interacts with is not changing anymore). But since it is not changing there is no INDUCED electric field inside the coils.. that induced electric field was opposing the electric field caused by the power source. But now it is gone, since someone stalled the motor. So no the net field is a bit higher and the current is indeed higher.

BUT, the fact that a thing suddenly is getting hotter does NOT necessarily mean that the current is larger..
Note that in this particular case, TWO things are causing the same effect!

And I still hope to hear about work and the sign of the work. You seem to have abandoned that aspect of the posts!
 
  • #15
Ott Rovgeisha said:
Electric resistance is more like simply ALL OF THE circumstances together
This is what I read and it seems to imply what you meant.

What else, apart for resistance, can make the motor get hotter (ignore the difference in cooling.)?
 
  • #16
sophiecentaur said:
This is what I read and it seems to imply what you meant.

What else, apart for resistance, can make the motor get hotter (ignore the difference in cooling.)?

Follow the logic:

1. If the resistance in a circuit is high, that means that the current is smaller.
2. If the current is smaller, that is equal to saying that there are fewer charge carriers going through a cross-sectional slice of the wire PER unit of time (supposing it is a metal wire)
3. Since there are fewer charge carriers accelerating through every slice, that means that there are less "collisions" with the atomic cores per unit time.
4. That means that the energy conversion from electronic kinetic energy into the kinetic energy of the atoms is going on way more slowly: in other words, the heating is way slower than it would be had the resistance of the circuit been smaller.
5. If the wire is getting hotter more slowly, then it also gets less hotter.. That is: the equilibrium temperature of heating and cooling is low...
6. That means that means that the wire remains cooler the higher the resistance of the circuit!
A motor essentially has wires coiled up as you know (way better than me).

To put it in other words: the higher the resistance the smaller the current, the smaller the current the smaller the power of the circuit, that means it gets heated more slowly and therefore it remains cooler. This power argument holds, however only if we do not make the wire thicker (but that is sort of a different story).

By power, I mean the rate at which the kinetic energy of an electron is turned into other forms. Note that HEAT is included in there.
Another story is that you may not be interested in heat, but in rotation.
But even rotation is caused by the electrons: in the coil, there is a magnetic field due to the magnets near the motor. That magnetic field CHANGES the direction of the electrons that are accelerating due to the electric field in the wire. There is an enormous amount of free electrons all abruptly changing their direction because of the magnetic field. They ALL at once collide with the atoms that make up the windings, and drag the atoms with them: this makes the motor turn.So to answer directly to your question: "what else apart from the resistance causes the wire to get hot...?"

The resistance on itself does not cause the heat at all. Heat is caused by the collisions of the free electrons with the atomic cores. (that is of course a simplification, not involving quantum mechanics, but it is not wrong, because we can define the word "collision" in such a manner).
You might say that the nature of atoms are part of creating the resistance to the current, but that does not mean that it directly causes the heat..

Consider this: if you have a thick wire connected to the thin wire, which is in turn, connected to a thick wire from the other end and you connect it with a voltage source you will note a particular thing: the thinner part that poses the most resistance to the whole current get hotter, but that is because in the thin part the electrons actually speed up MORE that in the wire, where there is less resistance to a current!
This is because the electric field is stronger in the thin part.
BUT had the thin wire been more thicker, it would have less resistance to a current and it would get EVEN MORE hotter due to an overall larger current.

While you might say that the part, what causes the most resistance to the current is getting more heated up, you must be very careful, because the resistance is a factor that makes the current smaller everywhere in the circuit. Smaller current, however automatically means less power.

There are so many subtle things here that it is a challenge to explain it in detail. This is the part of the reason that a sizable number of scientists just use equations to "explain" something and saying that physics is all math. I cannot agree with it. It is a dangerous idea. Math helps physics.. helps to add some ideas where the words might end, but it should not replace it or become a shield for the lack of understanding of concepts. Concepts can be horribly non-intuitive, but still, that does not matter.And finally: of course ordinary friction can make and does make the motor hotter, in fact a huge part of the energy goes into the heating due to the friction.
It is just like pushing a sledge 10 km with a uniform speed: note that ALL of the energy, except for the little kick in the beginning of the pushing, goes into heat. All of it. The only energy that went into speed was the very brief acceleration of the sledge at the beginning of the pushing.
 
  • #17
I have one comment to make about the above post and that is Resistance is merely the constant of proportionality between Potential Difference and Current. It's only a derived quantity. The current that flows in a component, as a result of the PD is governed by a number of mechanisms which are not just Ohmic 'Resistance'. When we talk of wires getting hot and resistance, that is only the most familiar example of Energy Transfer. V and I may be related due to EM (radio antennae etc.) and Electromechanical examples (motors). For a finite value of Resistance to be measured, all that's necessary is for Energy to be transferred. Where there is Loss, the 'direction' of flow of Energy can be only one way, which is what this thread is basically about, I think. We're talking Entropy, basically.
 

What is the coefficient of static friction?

The coefficient of static friction is a measure of the amount of force required to prevent an object from moving on a surface. It is a dimensionless quantity and is denoted by the symbol "μ".

How is the coefficient of static friction calculated?

The coefficient of static friction can be calculated by dividing the maximum force required to keep an object from sliding by the normal force acting on the object. This can be represented by the equation μ = F/N, where μ is the coefficient of static friction, F is the maximum force, and N is the normal force.

Why is calculating the coefficient of static friction important?

Calculating the coefficient of static friction is important because it helps determine the force needed to keep an object from sliding on a surface. This information is essential for designing and building structures, as well as ensuring the safety of objects and people in various situations.

What factors affect the coefficient of static friction?

The coefficient of static friction can be affected by various factors such as the type of surface, the roughness of the surface, and the weight of the object. It is also dependent on the materials in contact and the presence of any lubricants between them.

What is the difference between coefficient of static friction and coefficient of kinetic friction?

The coefficient of static friction is the measure of the maximum force required to keep an object from sliding on a surface. On the other hand, the coefficient of kinetic friction is the measure of the force required to maintain a constant velocity of an object on a surface. The coefficient of kinetic friction is usually lower than the coefficient of static friction.

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