Spring acting on an object (dynamics)

[Femme_physics]

My frenzy is not over yet! :)

Homework Statement

http://img607.imageshack.us/img607/5116/boingyboingy.jpg

A box whose mass is 12 kg is at rest on a horizontal plane, and pressed by a spring as described. The spring, pressed from its standard condition at point A till point O, across a length of 0.25m, and locked by a brake. AFter releasing the box from rest at point O, calculate:

A) The velocity of the box at point B (OB = 0.5m)
B) The distance s that the box will cross from point O till it stops (at point D).

Given:
Spring hardness coefficient = 800 N/m
Kinetic friction between the box and the ground = 0.15

Comment: an elastic work force on a spring on runway s (W) equals

http://img838.imageshack.us/img838/9175/wwwwwwp.jpg

The Attempt at a Solution

Once again I'm getting sigma F = ma NOT EQUAL!

Also, my V (or Vb) appears to be incorrect... help, please?


http://img163.imageshack.us/img163/9843/spring1i.jpg

http://img31.imageshack.us/img31/8172/spring2k.jpg

http://img833.imageshack.us/img833/5654/spring3b.jpg

 

[I like Serena]

Couldn't you have a frenzy on a day that I'm free? :P

 

[Femme_physics]

You've been amazing so far, you have real life and you don't have to overreach! :) Maybe other helpers will get to me if you won't. Otherwise ths problem will have to wait till tomorrow since I only have 3 hours. But to answer your rhetorical question. No, I can't control the timing of my frenzies :( It's like an eating binge, it just hits you all of a sudden! Not...that... I have any eating binges... are you calling me fat?

 

[I like Serena]

This is truly an energy formula problem.
You could solve it with force formulas, but that requires calculus (I think).

To do this problem asks for a bit more knowledge on energy though.
I'll try to explain.
I'll leave it up to you what you want to do with it. First I'll give one more formula:

Kinetic energy E_k = m v² / 2The compressed spring contains an energy given by the problem (W = c x s² / 2).
This energy is fully converted into kinetic energy of the box on the trajectory O-->A, minus the work done by friction.
After that the box is slowed down by the friction force until it stops.

At point A, the box will have gained the energy from the spring, which is:
E_el = c x OA² / 2

And it will have lost energy equal to the work done by friction, which is:
W_friction = Fs x OA

So at point A, the box would have a kinetic energy
E_k = c x OA² / 2 - Fs x OA

With the formula for kinetic energy you can calculate the speed at A...

 

[Femme_physics]

Wow.

Erm,

Okay...let me try and break it step by step :)

c x OA² / 2

So,

800 x 0.25² / 2

The result of this is 25.
25 "energy the box has gained", I presume. The units are Joules since this is energy, yes?

Where did this formula come from, btw? I'm looking at my formulas sheet (this is the sheet my lecturer gave to the entire class) and I just can't see it.

http://www.scribd.com/doc/57839921/All-Dynamics-Formulae

 

[I like Serena]

Wow.

Erm,

Okay...let me try and break it step by step :)

c x OA² / 2

So,

800 x 0.25² / 2

The result of this is 25.
25 "energy the box has gained", I presume. The units are Joules since this is energy, yes?

Yes!

Where did this formula come from, btw? I'm looking at my formulas sheet (this is the sheet my lecturer gave to the entire class) and I just can't see it.

http://www.scribd.com/doc/57839921/All-Dynamics-Formulae

On page 5 you have:

$E_{el} = \frac 1 2 k x^2 \qquad$ :אנרגיה פוטנציאלית אלסטית

That's the one!

 

[Femme_physics]

And it will have lost energy equal to the work done by friction, which is:
W_friction = Fs x OA

Well,

Sum of all forces on X = -W + N = 0
N = W = 12(9.81) = 117.72 [N]
Fs = N x Ms
Fs = 117.72 x 0.15 = 17.66 [N]

W_friction = 17.66 x 0.25 = 4.41 [Joules]

So E_k is in fact the same thing as the formula you wrote to me here


Now I don't see what formula you've used to convert our 2 results with the unit of Joules to this:

So at point A, the box would have a kinetic energy
Ek = c x OA2 / 2 - Fs x OA

From this:

E_k = m v² / 2

I don't see the mass in here, and I don't see velocity in here.

 

[I like Serena]

Well,

Sum of all forces on X = -W + N = 0
N = W = 12(9.81) = 117.72 [N]
Fs = N x Ms
Fs = 117.72 x 0.15 = 17.66 [N]

W_friction = 17.66 x 0.25 = 4.41 [Joules]

Good!

So E_k is in fact the same thing as the formula you wrote to me here


Now I don't see what formula you've used to convert our 2 results with the unit of Joules to this:



From this:

E_k = m v² / 2

I don't see the mass in here, and I don't see velocity in here.

No, but they are equal to each other, so:

m v_A² / 2 = c x OA2 / 2 - Fs x OA

From this you can solve v_A!

Energy is something that is converted from one form to another.
The elastic energy contained in the compressed spring is converted to kinetic energy of the block.
And while the block is sliding to the right, it bleeds energy in the form of heat (friction).

 

[Femme_physics]

Let me see if I understand the formula.

Kinetic energy is equal to gained divided by lost potential energy?

 

[I like Serena]

Let me see if I understand the formula.

Kinetic energy is equal to gained divided by lost potential energy?

Huh?

I'd say: kinetic energy gained by the box is equal to the elastic energy lost by the spring.

There is no division in there.

And btw, some of the energy is bled away by friction, which is subtracted (not divided!).

 

[Femme_physics]

Well, my pre-finals test is tomorrow so I think I'll skip the energy part for now, but will definitely go over it before the finals to feel comfortable with it! Thanks^^

 

[Femme_physics]

I'm finding it a bit complex to use those formulas step by step, nor can I really keep track. Can I solve it using this?:http://img845.imageshack.us/img845/616/mgh.jpg [/QUOTE]

 

[I like Serena]

I'm finding it a bit complex to use those formulas step by step, nor can I really keep track. Can I solve it using this?

Yes and no.
This is the energy conservation formula for gravity energy, kinetic energy and elastic energy.
(I'm avoiding the word "potential energy" here because that would be ambiguous.)
It's very useful, except that it does not account for friction.

But let's use this formula for now, just to get started.
It will be good practice, help your understanding and you'll probably be using it to solve other problems.

That is, let's assume for now that there is no friction.
Can you apply this formula then and tell me what the speed of the box at point A will be?

 

[Femme_physics]

Huh?

I'd say: kinetic energy gained by the box is equal to the elastic energy lost by the spring.

There is no division in there.

And btw, some of the energy is bled away by friction, which is subtracted (not divided!).

I finally realized what confused me this entire time! Sooooooooooo glad I pinned it.

I translated this:m v_A² / 2 = c x OA2 / 2 - Fs x OATo this:

http://img3.imageshack.us/img3/6553/baffledf.jpg You have to admit it's a valid confusion!

(anyway now that I finally see it I'll get cracking!)

 

[Femme_physics]

And here's my solution attempt for Va

http://img215.imageshack.us/img215/6595/mvab.jpg

it appears to be wrong according to the manual, drat

 

[I like Serena]

Aherm, good morning Femme_physics!

I finally realized what confused me this entire time! Sooooooooooo glad I pinned it.

Aha!

And here's my solution attempt for Va

it appears to be wrong according to the manual, drat

What formula are you using here?
There seems to be a couple of things wrong with it - division instead of subtraction for one...

 

[Femme_physics]

Oh, damn it, I dragged my confusion from before lol, oops!

there!

http://img651.imageshack.us/img651/9716/13mmd.jpg Closer this time, but still slightly off. You said I had a "couple of things wrong"...couple. Where's the couple? I want to meet them. I dislike them already! We're so not double-dating with them.

 

[I like Serena]

Oh, damn it, I dragged my confusion from before lol, oops!

Aherm, good morning Femme_physics!

there!

Closer this time, but still slightly off. You said I had a "couple of things wrong"...couple. Where's the couple? I want to meet them. I dislike them already! We're so not double-dating with them.

What formula for kinetic energy are you using here?
It kind of looks like a double-date with centripetal force.

 

[Femme_physics]

Is that it?

http://img717.imageshack.us/img717/3645/mvar.jpg

By the by, this is what the solution manual did if you can even see it(?) *sighs--I miss my scanner*

http://img221.imageshack.us/img221/1244/hopemv.jpg

It only appears to give the result for Vb.

 

[I like Serena]

Aherm, good morning Femme_physics!
Aherm, good morning Femme_physics!
Aherm, good morning Femme_physics!

Is that it?

Huh?
That's the same wrong formula with the couple of mistakes as before.
So you do want to double-date with them?

By the by, this is what the solution manual did if you can even see it(?) *sighs--I miss my scanner*

It only appears to give the result for Vb.

Ah, yes. In the problem they ask for the speed at B.
The method is the same, but you need to plug in the distance to B in the formula for the work done by friction.

 

[Femme_physics]

LOL sorry getting overindulgent with the exercise; Good morning dear I like Serena! :)

That's the same wrong formula with the couple of mistakes as before.

Ah I linked the wrong file oops!

http://img687.imageshack.us/img687/7926/mvaaaaa.jpg

The method is the same, but you need to plug in the distance to B in the formula for the work done by friction.

I see. Step by step then^^ :)

 

[I like Serena]

LOL sorry getting overindulgent with the exercise; Good morning dear I like Serena! :)

Thank you!

Ah I linked the wrong file oops!

I see. Step by step then^^ :)

Very good!

So do you know what to do next now?

 

[Femme_physics]

I can throw a guess at best...

http://img807.imageshack.us/img807/6067/mvbu.jpg

 

[I like Serena]

Good guess!
But you filled in a distance of 0.5 meters for the work done by friction.
What is the distance from A to B?
Btw, as an alternative you can simply repeat your original calculation, plugging in OB instead of OA.

 

[Femme_physics]

You're right! I was wondering that. So all I should've written for distance is 0.25 and I had it right?

 

[I like Serena]

You're right! I was wondering that. So all I should've written for distance is 0.25 and I had it right?

Yes.

 

[Femme_physics]

Awesome! :) I feel...so... empowered!

Thanks, ILS!

 

[I like Serena]

Awesome! :) I feel...so... empowered!

Thanks, ILS!

I like to empower you!

So are you done now with this problem?
Because I thought there was more ...

 

[Femme_physics]

But that means I'll be making "more" mistakes :(

Oh, wait, I got you tagging along for the ride :)

http://img716.imageshack.us/img716/8371/ssssovv.jpg

 

[I like Serena]

But that means I'll be making "more" mistakes :(

Oh, wait, I got you tagging along for the ride :)

Yes, more mistakes.
I see another couple again.
Wanna double-date?

Ok, here we go.

In your formula you write s for the distance that friction "works".
But that is not s, but only the distance from A to C, that is, AC.
s is a longer distance, starting from O.

And then you have $\frac {m v_A^2} 2$ in your formula, which is good!
But you filled in the numbers as if it was $m v_A$, which it isn't!

 

[Femme_physics]

You'll have to excuse my indiscretion. I'm not like that at tests! It's just that when I'm pretty sure I'm wrong about something, I hardly bother to double-triple check myself.

Now okay, back to the exercise!

Starting with this...

But that is not s, but only the distance from A to C, that is, AC.

I don't see point C on the diagram

 

[I like Serena]

You'll have to excuse my indiscretion. I'm not like that at tests! It's just that when I'm pretty sure I'm wrong about something, I hardly bother to double-triple check myself.

Now okay, back to the exercise!

Starting with this...

I don't see point C on the diagram

Very good! I was just checking if you're still sharp! *cough*

 

[Femme_physics]

ROFL!

You're getting funnier and funnier I see

So what did you mean instead of C?

 

[I like Serena]

So what did you mean instead of C?

Oh, point D.
What did you have on the left hand side of your last formula?
I read it as $\frac {m v_C^2} 2$, but it should be $\frac {m v_D^2} 2$.

 

[Femme_physics]

So from O to A when you release the sprint there's no friction playing part? That's impossible! We know that friction always acts. How can it be?