Spring acting on an object (dynamics)

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The discussion revolves around a physics problem involving a 12 kg box pressed by a spring, where participants seek to calculate the box's velocity and stopping distance after being released. Key points include the application of energy conservation principles, where the elastic potential energy from the spring is converted into kinetic energy, while also accounting for work done against friction. Participants express confusion over the formulas and calculations, particularly regarding how to properly incorporate friction and the correct distances in their equations. There is a collaborative effort to clarify these concepts, with references to specific energy formulas and the importance of understanding the relationship between kinetic and elastic energy. The conversation highlights the complexity of dynamics problems and the necessity of careful step-by-step problem-solving.
Femme_physics
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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
 
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Couldn't you have a frenzy on a day that I'm free? :P
 
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?
 
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. :smile:First I'll give one more formula:

Kinetic energy Ek = 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:
Eel = c x OA2 / 2

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

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

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

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
 
Femme_physics said:
O_o

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! :smile:


Femme_physics said:
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! :wink:
 
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And it will have lost energy equal to the work done by friction, which is:
Wfriction = 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]

Wfriction = 17.66 x 0.25 = 4.41 [Joules]

So Ek 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:

Ek = m v2 / 2

I don't see the mass in here, and I don't see velocity in here. :confused:
 
Femme_physics said:
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]

Wfriction = 17.66 x 0.25 = 4.41 [Joules]

Good! :smile:


Femme_physics said:
So Ek 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:

Ek = m v2 / 2

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

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

m vA2 / 2 = c x OA2 / 2 - Fs x OA

From this you can solve vA!

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).
 
Let me see if I understand the formula.

Kinetic energy is equal to gained divided by lost potential energy?
 
  • #10
Femme_physics said:
Let me see if I understand the formula.

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

Huh? :rolleyes:

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!).
 
  • #11
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^^
 
  • #12
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  • #13
Femme_physics said:
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?
 
  • #14
I like Serena said:
Huh? :rolleyes:

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 vA2 / 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!)
 
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  • #15
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  • #16
Aherm, good morning Femme_physics! :smile:
Femme_physics said:
I finally realized what confused me this entire time! Sooooooooooo glad I pinned it.

Aha!
Femme_physics said:
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... :wink:
 
  • #17
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.
 
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  • #18
Femme_physics said:
Oh, damn it, I dragged my confusion from before lol, oops!

Aherm, good morning Femme_physics! :smile:


Femme_physics said:
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. :wink:
 
  • #20
Aherm, good morning Femme_physics! :smile:
Aherm, good morning Femme_physics! :smile:
Aherm, good morning Femme_physics! :smile:

Femme_physics said:
Is that it?

Huh? :rolleyes:
That's the same wrong formula with the couple of mistakes as before.
So you do want to double-date with them? :rolleyes:
Femme_physics said:
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.
 
  • #21
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^^ :)
 
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  • #22
Femme_physics said:
LOL sorry getting overindulgent with the exercise; Good morning dear I like Serena! :)

Thank you! :smile:


Femme_physics said:
Ah I linked the wrong file oops!

I see. Step by step then^^ :)

Very good! :smile:

So do you know what to do next now?
 
  • #24
Good guess! :smile:
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.
 
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  • #25
You're right! I was wondering that. So all I should've written for distance is 0.25 and I had it right?
 
  • #26
Femme_physics said:
You're right! I was wondering that. So all I should've written for distance is 0.25 and I had it right?

Yes.
 
  • #27
Awesome! :) I feel...so... empowered!

Thanks, ILS!
 
  • #28
Femme_physics said:
Awesome! :) I feel...so... empowered!

Thanks, ILS!

I like to empower you! :blushing:

So are you done now with this problem?
Because I thought there was more ... :wink:
 
  • #29
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  • #30
Femme_physics said:
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! :smile:
But you filled in the numbers as if it was m v_A, which it isn't! :rolleyes:
 
  • #31
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
 
  • #32
Femme_physics said:
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*
 
  • #33
ROFL!

You're getting funnier and funnier I see :smile:

So what did you mean instead of C?
 
  • #34
Femme_physics said:
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.
 
  • #35
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?
 
  • #36
Femme_physics said:
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?

Oh yes, there is friction in OA, but when you get to A and have a speed vA, that friction has already done its work resulting in the speed vA.
After that you should not apply that friction a second time! :wink:
 
  • #38
Femme_physics said:
Nailed it! :D

Or...erm...sprung it... *shrugs* *scratches head*

Got it :)

Thanks ;)

Let's nail it down before it springs away again! :D
 
  • #39
I solved a similar, even tougher problem and posted it here just so you'd be proud of me :smile:

To show you that the knowledge you imbue in me really sticks!

Homework Statement



http://img827.imageshack.us/img827/1787/compartment1.jpg

In the drawing is described a product line of product M. When the compartment opens, one product falls on the horizontal table. At that moment is released a compressed spring K, that gives product M a push, and the product moves towards the segment AB on the horizontal table (see drawing). Given:

Friction coeffecient between product and desk = 0.1
Spring's constant: k = 2N/cm
Length of the spring at relaxe position: L1 = 200mm
Length of spring at compressed position: L2 = 140mm

Calculate the velocity of the product at point B (Vb). The measurements in the drawing are given in mm.

Comments:

A) Potential energy E stored in the spring in a compressed state equals:

http://img849.imageshack.us/img849/3199/compartment2.jpg

B) In calculate speed Vb do not mind the falling of the product form the compartment, only its motion across the desk (between point A and B)


(*The distance AB = 250mm)
http://img838.imageshack.us/img838/5246/compartment3.jpg

AND HERE IS THE SOLUTION I JUST COMPLETED

http://img200.imageshack.us/img200/4513/vbfinal.jpg
 
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  • #40
Femme_physics said:
I solved a similar, even tougher problem and posted it here just so you'd be proud of me :smile:

To show you that the knowledge you imbue in me really sticks!

I'm proud of you! :smile:

(Did you solve the problem while typing out a question? :wink:)
 
  • #41
I like Serena said:
I'm proud of you! :smile:

(Did you solve the problem while typing out a question? :wink:)

LMAO yes, you're on to me ;) but I did fail it, I just noticed my error as I was typing it. I doubt I could've solved it on my own since I'm so new to energy-work-spring problems which is why I immediately go to the council of geniuses (i.e. physics forums). But when it comes to 3D static stuff I no longer jump to post here. :-) I always pick up my own mistakes.
 
  • #42
Now that I see Fk it looks like the F word! :wink:

And working with energy-work is confusing when you start doing it, but soon you'll find it's much easier than force and moment sums. :smile:
 
  • #43
Still a bit confusing!

Once again a bit struggling with an elastic energy problem

In the drawing is described a product line of product M. When the compartment opens, drops a product on the horizontal desk. At that moment a compressed spring K is released giving the product a push, and it proceeds along AB of the horizontal desk (see drawing)

Given:

Product weight G = 5 [N]
Friction coeffecient between desk and product = 0.1
Spring's constant: c = 2N/cm
Length of spring at relaxed position: L1 = 180mm
Length of spring in compressed positipn: L2 = 130

Calculate Vb. Measurements are in the drawing given in mm. In calculating the velocity, ignore the drop of the product from the compartment, only take in account its movement across the desk)
http://img59.imageshack.us/img59/9734/commentspringstuff.jpg

-----------------------------------------------------------------------http://img857.imageshack.us/img857/9369/elasticenergy.jpg

The answer is supposed to be 0.7 m/s
 
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  • #44
did u solve the problem
 
  • #45
i don't know english and i couldn't understand did u solve this problem or not
 
  • #46
So you say the box moves at point A, when the spring is still completely compressed?
I say there's no movement yet, no work against friction. So what do we have at point A?
Remember the box is pushed by the spring some before sliding alone. How long is this sliding path and where does it start?

Try to think of this system in motion, what parts move, by how much and when do they stop affecting each other? Mark those points on the drawing.
 
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  • #47
Hi Fp! :smile:

You have introduced an extra point between A and B where the spring would be relaxed, and then you called this point A.
But this new point isn't A.
Point A is the point where the spring is compressed.

Actually, there is no need to introduce this extra point.
But if you do, you need to correct your distances. ;)
 
  • #48
I got the answer here :smile: Didn't have time to scan the paper, but I did what you said and used the right distance and got the same answer as the manual. Thank you!
 
  • #50
You're brightening up my day! :smile:
 

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