Spring acting on an object (dynamics)

In summary, the problem involves a box with a mass of 12 kg on a horizontal plane, pressed by a spring and released at point O. The goal is to calculate the velocity at point B (OB = 0.5m) and the distance the box will travel from point O until it stops at point D. Given the spring hardness coefficient of 800 N/m and a kinetic friction of 0.15 between the box and the ground, the solution involves calculating the kinetic energy gained by the box from the compressed spring and subtracting the energy lost from friction. The formula for kinetic energy is Ek = m v^2 / 2, where m is the mass and v is the velocity. The energy lost from friction is
  • #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:
 
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  • #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:
 
  • #51
It's darkening my day because I got it wrong!

Since he already gave out the answer, I wanted to ask why I am wrong. I got the first question wrong embarrassingly

i)

[tex]\int_{x_0 = -0.25m}^{x = 0.25} -kx \; dx - \mu mg \Delta x = \Delta K[/tex]

[tex]- \mu mg \Delta x = \frac{1}{2}mv^2[/tex]

Basically I couldn't take the negative square root since [tex]\Delta x > 0[/tex] and when I do break some rules I get the wrong answer.
 
  • #52
Hmm, you made the force of the spring cancel against itself...? :confused:
 
  • #53
I like Serena said:
Hmm, you made the force of the spring cancel against itself...? :confused:

It arrives at point B? A is the equilibrium point?
 
  • #54
Assuming you're talking about the last problem...

A is not the equilibrium point.

The problem states:
Length of the spring at relaxed position: L1 = 200mm
Length of spring at compressed position: L2 = 140mm

Point A is the point where the spring is compressed.
Then the force of the spring is going to act from the compressed position up to the relaxed position.
After that it looses contact with the box, and the spring does not keep acting pulling the box back again.
As it is, you have integrated with the wrong boundaries, and you also integrated in such a way that the transferred energy from the spring is zero, which it isn't.
 
  • #55
There is a last problem? Lol

I am just looking at

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
 
  • #56
flyingpig said:
There is a last problem? Lol

I am just looking at

Right! :smile:
So forget what I said about the length of the spring.
And so yes, A is the equilibrium point.

What remains is that you integrated from -0.25 m to 0.25 m, while you should have been integrating from -0.25 m to 0 m, where the spring loses contact with the box (and does not pull it back).
 
  • #57
why doesn't the spring extend to x = 0.25m? How would I know the spring loses contact at the equilibrium point?
 
  • #58
flyingpig said:
why doesn't the spring extend to x = 0.25m? How would I know the spring loses contact at the equilibrium point?

The spring loses contact, because it is not attached to the box.
If it were it would pull the box back.

The spring does not extend to x = 0.25 m, since it has no energy left, having transferred all its energy to the box.
 
  • #59
Is that true with all springs? That they all lose contact at the equilibrium point? Does that mean x_final < 0 always?
 
  • #60
It's true for "ideal" springs - it has to.
And more specifically x_final = 0 always (assuming the spring is not attached of course).
Otherwise, the spring would adjust infinitely fast since it has no mass.

Now, real-life springs, that's quite another matter.
Shall we put them aside until a more advanced course?
 
  • #61
Why do ideal springs stop at xf = 0? We have friction in this problem, does that mean it really doesn't stop at xf = 0, but xf < 0?
 
  • #62
Femme you have a lot of interesting physics problems. I want your book lol
 
  • #63
flyingpig said:
Why do ideal springs stop at xf = 0? We have friction in this problem, does that mean it really doesn't stop at xf = 0, but xf < 0?

Yes, you're right. :)
If the block is stopped by friction before xf = 0, then we will have xf < 0.
That's what you get for things not being ideal (since the problem is not "frictionless")! ;)
 
  • #64
Is that also true for vertical springs?
 

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