Spring Constant Calculation for a 0.2kg Bloc with 40cm Displacement

[astrololo]

Homework Statement

A bloc of 0,2 kg is attached on a horizontal spring without any friction on the floor. The other side of the sprin gis attached to a wall. He does a movement of back and forth . Between the two extremities of the back and forth movement, the bloc has a displacement of 40 cm. The maximum speed of the bloc while he is oscillating is 0,5 m/s. What is the constant of the spring ?

Homework Equations

Kf=1/2*m*(vf)^2

The Attempt at a Solution

So we know that the maximum speed is 0,5 m/s. Then the kinetic energy is going to be kf= 1/2*0,2 kg* 0,5^2 = 0,025 J

Other than that, I'm not sure where to go. I know that my speed at the two extremities are going to be 0 m/s...

 

[PeroK]

Homework Statement

A bloc of 0,2 kg is attached on a horizontal spring without any friction on the floor. The other side of the sprin gis attached to a wall. He does a movement of back and forth . Between the two extremities of the back and forth movement, the bloc has a displacement of 40 cm. The maximum speed of the bloc while he is oscillating is 0,5 m/s. What is the constant of the spring ?

Homework Equations

Kf=1/2*m*(vf)^2

The Attempt at a Solution

So we know that the maximum speed is 0,5 m/s. Then the kinetic energy is going to be kf= 1/2*0,2 kg* 0,5^2 = 0,025 J

Other than that, I'm not sure where to go. I know that my speed at the two extremities are going to be 0 m/s...

If the speed is 0, where has all the kinetic energy gone?

 

[astrololo]

If the speed is 0, where has all the kinetic energy gone?

To the potential energy.

 

[PeroK]

To the potential energy.

Correct. And what do you know about the potential energy of a spring?

 

[astrololo]

Correct. And what do you know about the potential energy of a spring?

This and this :

U=1/2*k*e^2

Kf+Uf=Ki+Ui

 

[gneill]

The Attempt at a Solution

So we know that the maximum speed is 0,5 m/s. Then the kinetic energy is going to be kf= 1/2*0,2 kg* 0,5^2 = 0,025 J

Other than that, I'm not sure where to go. I know that my speed at the two extremities are going to be 0 m/s...

So, where might the KE have gone (where's it hiding) when the block is at one of the extremities?

 

[astrololo]

So, where might the KE have gone when the block is at one of the extremities?

It became a potential energy. This :
U=1/2*k*e^2
Kf+Uf=Ki+Ui

 

[gneill]

It became a potential energy. This :
U=1/2*k*e^2
Kf+Uf=Ki+Ui

Right. Continue...

 

[astrololo]

Right. Continue...

So :

Kf+Uf=Ki+Ui

0,025 J + 1/2*k*e^2 = 0 J + 1/2*k*e^2

 

[gneill]

So :

Kf+Uf=Ki+Ui

0,025 J + 1/2*k*e^2 = 0 J + 1/2*k*e^2

Where is the block when its speed is maximum? Where is the block when its speed is zero?

 

[astrololo]

Where is the block when its speed is maximum? Where is the block when its speed is zero?

Well, when the speed is 0 its going to be at the two extremities. When its at the maximum its going to be somewhere in the middle.

 

[gneill]

Well, when the speed is 0 its going to be at the two extremities.

Right.

When its at the maximum its going to be somewhere in the middle.

Not just somewhere: There's a precise location. You should be able to deduce it from the energy equation. When KE is maximum, what is the PE?

 

[astrololo]

Right.

Not just somewhere: There's a precise location. You should be able to deduce it from the energy equation. When KE is maximum, what is the PE?

Oh, there is going to be a potential energy of 0 because it will have been transferred all to the kenetic one right ?

 

[gneill]

Oh, there is going to be a potential energy of 0 because it will have been transferred all to the kenetic one right ?

Right. So what can you say about the values of maximum KE and maximum PE? What are the values of displacement associated with each?

 

[astrololo]

Right. So what can you say about the values of maximum KE and maximum PE? What are the values of displacement associated with each?

So the maximum is going to be 0,025 for the kinetic one and 0,025 for the potential one at the two extremities.

 

[gneill]

So the maximum is going to be 0,025 for the kinetic one and 0,025 for the potential one at the two extremities.

Assuming that you meant Joules for the units, yes. And you should be able to locate the displacements involved from the given information. Then just write out your energy equation accordingly.

 

[astrololo]

Assuming that you meant Joules for the units, yes. And you should be able to locate the displacements involved from the given information. Then just write out your energy equation accordingly.

I'll go try this and give you my thanks if it works !

 

[astrololo]

Assuming that you meant Joules for the units, yes. And you should be able to locate the displacements involved from the given information. Then just write out your energy equation accordingly.

Ok so I obtained : 0,025 J + 0.5*k*e^2 = 0 J + 0,025 J

 

[gneill]

Ok so I obtained : 0,025 J + 0.5*k*e^2 = 0 J + 0,025 J

First state what the conditions are for each side of the equation. Numbers alone aren't telling the whole tale. It doesn't help to have your known maximum numerical value on both sides of the equation at the same time, since they'll just cancel out.

Remember that when KE is maximum the PE is zero, and when PE is maximum the KE is zero. Choose your two block positions so that you can use your maximum energy value to best advantage.

 

[astrololo]

First state what the conditions are for each side of the equation. Numbers alone aren't telling the whole tale. It doesn't help to have your known maximum numerical value on both sides of the equation at the same time, since they'll just cancel out.

Remember that when KE is maximum the PE is zero, and when PE is maximum the KE is zero. Choose your two block positions so that you can use your maximum energy value to best advantage.

This the equation ; Kinetic energy final + potential energy final = kinetic energy initial + potential energy initial

I thought that I should do the following :

Kinetic energy final : 0,025 J

Potential energy final : 0 J

Kinetic energy initial : 0 J

Potential energy final : 0,025 J

 

[astrololo]

First state what the conditions are for each side of the equation. Numbers alone aren't telling the whole tale. It doesn't help to have your known maximum numerical value on both sides of the equation at the same time, since they'll just cancel out.

Remember that when KE is maximum the PE is zero, and when PE is maximum the KE is zero. Choose your two block positions so that you can use your maximum energy value to best advantage.

It's really the position which is bugging me... I don't knwo what to do with 40 cm...

 

[gneill]

This the equation ; Kinetic energy final + potential energy final = kinetic energy initial + potential energy initial

I thought that I should do the following :

Kinetic energy final : 0,025 J

Potential energy final : 0 J

Kinetic energy initial : 0 J

Potential energy final : 0,025 J

Presumably that last entry should be "Potential energy initial"?
If the KE final is 0.025 J, that must mean that the final position is at the midpoint. And your initial position is at one of the maximum displacement points.

So when you write your equation, don't use a number for the maximum displacement, use the expression that includes the displacement. What is the maximum displacement? ...

It's really the position which is bugging me... I don't knwo what to do with 40 cm...

Doesn't that tell you what the maximum displacement should be?

 

[astrololo]

Presumably that last entry should be "Potential energy initial"?
If the KE final is 0.025 J, that must mean that the final position is at the midpoint. And your initial position is at one of the maximum displacement points.

So when you write your equation, don't use a number for the maximum displacement, use the expression that includes the displacement. What is the maximum displacement? ...Doesn't that tell you what the maximum displacement should be?

I think I am going to cry. Why does the maximum happen to be at the middle point and not at the third, fourth, etc.? I don't even know where my initial position is (Without contracting or lenghting my spring)

And yeah I meant initial its a mistake

 

[gneill]

In an oscillator like a spring/mass system, energy is traded back and forth between kinetic energy of the mass and potential energy of the spring. The kinetic energy depends upon the speed of the mass, while the potential energy depends upon the stretch or compression (displacement) of the spring from its equilibrium position. In this case the equilibrium position corresponds to the location where the spring is relaxed: zero displacement.

When kinetic energy is maximum the potential energy is minimum, i.e. zero. Zero potential energy means the displacement is zero, since PE = (1/2) k 0². That places the maximum KE at the midpoint where displacement from equilibrium is zero. The mass zips through the midpoint at its maximum speed, on its way to the next maximum displacement.

When potential energy is maximum the kinetic energy is minimum, i.e. zero. Zero kinetic energy occurs when the spring is maximally stretched (the mass is at maximum displacement and has zero speed). Then the potential energy is PE = (1/2) k Δx_max², where Δx_max is the maximum displacement value.

You have been given the maximum speed and the total displacement. So you can calculate both the maximum KE (which you've done) and the maximum displacement (which you haven't done yet).

 

[astrololo]

In an oscillator like a spring/mass system, energy is traded back and forth between kinetic energy of the mass and potential energy of the spring. The kinetic energy depends upon the speed of the mass, while the potential energy depends upon the stretch or compression (displacement) of the spring from its equilibrium position. In this case the equilibrium position corresponds to the location where the spring is relaxed: zero displacement.

When kinetic energy is maximum the potential energy is minimum, i.e. zero. Zero potential energy means the displacement is zero, since PE = (1/2) k 0². That places the maximum KE at the midpoint where displacement from equilibrium is zero. The mass zips through the midpoint at its maximum speed, on its way to the next maximum displacement.

When potential energy is maximum the kinetic energy is minimum, i.e. zero. Zero kinetic energy occurs when the spring is maximally stretched (the mass is at maximum displacement and has zero speed). Then the potential energy is PE = (1/2) k Δx_max², where Δx_max is the maximum displacement value.

You have been given the maximum speed and the total displacement. So you can calculate both the maximum KE (which you've done) and the maximum displacement (which you haven't done yet).

Sorry, I guess I am tired today because its going through my head. I did this picture : http://imgur.com/gUvMPQz

I know that my displacement is e=L-Lnatural

Lnatural = 0 because this is the position where there isn't any force applied on it. Now, where is the 0 position on my picture ? I understand why my displacement is 0 because of the PE formula.

 

[gneill]

Sorry, I guess I am tired today because its going through my head. I did this picture : http://imgur.com/gUvMPQz

I know that my displacement is e=L-Lnatural

Lnatural = 0 because this is the position where there isn't any force applied on it. Now, where is the 0 position on my picture ? I understand why my displacement is 0 because of the PE formula.

Right in the center of course. The midpoint.

 

[astrololo]

Right in the center of course. The midpoint.

Ok, we're almost here. This is the only point which is blocking me form resolving the problem. IM REALLY SOSRRY if I am taking your time!

I understand the following : When my kinetic energy is at the maximum, the position of my block will be 0 because this is according to the potential energy formula. FINE ! Now, why would I put this 0 in the middle ? (Pls don't be angry at me)

Here is a picture updated : http://imgur.com/9GbF6Wt

 

[gneill]

It's in the middle because that's where the displacement from equilibrium (relaxed spring, no forces) is zero.

If the system had no kinetic energy and no potential energy either, then the whole thing would be at rest with the mass sitting in the middle, no compression or extension of the spring. Energy is added by some external mechanism (such as your hand) moving the block away from the equilibrium point.

 

[astrololo]

It's in the middle because that's where the displacement from equilibrium (relaxed spring, no forces) is zero.

If the system had no kinetic energy and no potential energy either, then the whole thing would be at rest with the mass sitting in the middle, no compression or extension of the spring. Energy is added by some external mechanism (such as your hand) moving the block away from the equilibrium point.

I found the answer of 1.25... So If I understood correctly, the natural position is always half of the the whole displacement right ? Whatever the situation ?

 

[gneill]

I found the answer of 1.25... So If I understood correctly, the natural position is always half of the the whole displacement right ? Whatever the situation ?

1.25 what? Always include units with results! Numerically it looks good, but if you submit a result to a marker without including the units you will be sure to lose marks!

In simple setups like simple harmonic motion the equilibrium position is in the center of the whole displacement. One could contrive situations where the spring constant changes over the range, but you're unlikely to come across such things in your present courses.

 

[astrololo]

1.25 what? Always include units with results! Numerically it looks good, but if you submit a result to a marker without including the units you will be sure to lose marks!

In simple setups like simple harmonic motion the equilibrium position is in the center of the whole displacement. One could contrive situations where the spring constant changes over the range, but you're unlikely to come across such things in your present courses.

OK thank you ! Last question before we end this : Why do you think I blocked on this particular point ? Why would this happen ? Lack of knowledge in physics ? I guess this is my best bet... Also, did I bother you?

 

[gneill]

OK thank you ! Last question before we end this : Why do you think I blocked on this particular point ? Why would this happen ? Lack of knowledge in physics ? I guess this is my best bet... Also, did I bother you?

We all come across concepts that, for whatever reason, seem to be counter intuitive at first and take some time and brain sweat to resolve. Once the "Aha!" moment occurs and you get it it's usually smooth sailing from there with a bit of practice. Learning to write energy conservation equations for different situations in order to uncover some particular parameter is one of those things that requires a bit of practice.

And no, you didn't bother me I volunteer my time because I enjoy helping students. Students who persevere and continue to make an effort when things get tricky are never a problem.

 

[astrololo]

We all come across concepts that, for whatever reason, seem to be counter intuitive at first and take some time and brain sweat to resolve. Once the "Aha!" moment occurs and you get it it's usually smooth sailing from there with a bit of practice. Learning to write energy conservation equations for different situations in order to uncover some particular parameter is one of those things that requires a bit of practice.

And no, you didn't bother me I volunteer my time because I enjoy helping students. Students who persevere and continue to make an effort when things get tricky are never a problem.

To be honest, I didn't get the "Aha" moment because I just assumed that what you said is true. I mean, I did think of the possibility that the middle would be the "0" but I wasn't 100 % sure because there was no indication that this would be the case. In any case, the problems given are simple and of this type that I just did. I need to learn more physics in my free time. It's good to know there are people who are here to help others. Thank you!