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.
  • #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)

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

- \mu mg \Delta x = \frac{1}{2}mv^2

Basically I couldn't take the negative square root since \Delta x > 0 and when I do break some rules I get the wrong answer.
 
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  • #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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