Hooke's law does not apply when there is friction, right?

In summary, the question asks for guidance on how to write an equation involving friction force and initial velocity. The scenario involves an object being thrown into a spring with a known friction coefficient and spring constant. The goal is to find the initial and final velocities at equilibrium. The provided equation does not fully describe the system due to the position dependent friction coefficient.
  • #1
phase0
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Summary:: I am not sure about how can I write an appropriate equation to a question which include friction force and initial velocity.

I came across a question that I am not sure how to write an equation.In the question, there is an object of mass m that is thrown into spring at v initial velocity.As soon as the spring starts to compress, the friction force also starts to act and we know the friction coefficient and spring constant.So in this case, how can we find the initial velocity and the final velocity in equilibrium state? I have foun an equation but I am not sure about how can I do it in latex ##m\dfrac{d^{2}x}{dt^{2}}+c\dfrac{dx}{dt}+kx=F_{0}\cos Wt## if you have any idea please share with me

[Moderator's note: moved from a technical forum.]
 
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  • #2
From what you describe, there is no external driving force, therefore the right hand side must be zero.

Look up "damped harmonic oscillator." There is a lot out there. This will get you started.

The initial velocity is something that is chosen or given. There must be wording in the physical situation that says what is it is. The final velocity depends on what "final" means, i.e. the instant in time you choose to look at the motion.

As for the question that you asked in the title of this thread, the answer is "No, not right." Hooke's law says that, for small displacements, the elastic restoring force exerted by a spring is proportional to the displacement. This is true whether there is friction or not. The net force is another story.

On edit: Please post the question as was given to you. Then we can help you better.
 
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  • #3
kuruman said:
From what you describe, there is no external driving force, therefore the right hand side must be zero.

Look up "damped harmonic oscillator." There is a lot out there. This will get you started.

The initial velocity is something that is chosen or given. There must be wording in the physical situation that says what is it is. The final velocity depends on what "final" means, i.e. the instant in time you choose to look at the motion.

As for the question that you asked in the title of this thread, the answer is "No, not right." Hooke's law says that, for small displacements, the elastic restoring force exerted by a spring is proportional to the displacement. This is true whether there is friction or not. The net force is another story.

On edit: Please post the question as was given to you. Then we can help you better.
phase0 said:
Summary:: I am not sure about how can I write an appropriate equation to a question which include friction force and initial velocity.

I came across a question that I am not sure how to write an equation.In the question, there is an object of mass m that is thrown into spring at v initial velocity.As soon as the spring starts to compress, the friction force also starts to act and we know the friction coefficient and spring constant.So in this case, how can we find the initial velocity and the final velocity in equilibrium state? I have foun an equation but I am not sure about how can I do it in latex ##m\dfrac{d^{2}x}{dt^{2}}+c\dfrac{dx}{dt}+kx=F_{0}\cos Wt## if you have any idea please share with me

[Moderator's note: moved from a technical forum.]
phase0 said:
Summary:: I am not sure about how can I write an appropriate equation to a question which include friction force and initial velocity.

I came across a question that I am not sure how to write an equation.In the question, there is an object of mass m that is thrown into spring at v initial velocity.As soon as the spring starts to compress, the friction force also starts to act and we know the friction coefficient and spring constant.So in this case, how can we find the initial velocity and the final velocity in equilibrium state? I have foun an equation but I am not sure about how can I do it in latex ##m\dfrac{d^{2}x}{dt^{2}}+c\dfrac{dx}{dt}+kx=F_{0}\cos Wt## if you have any idea please share with me

[Moderator's note: moved from a technical forum.]
Oh,thank you!Here is my question,I just did not share the original version because I did not want to be understood as solving my question to you
 

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  • #4
I read the posted question and found the friction doesn't follow a simple, viscous-like, form. It states a position dependent friction coefficient. The posted equation, even with a null RHS, doesn't describe the system's behavior
 
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  • #5
Gordianus said:
I read the posted question and found the friction doesn't follow a simple, viscous-like, form. It states a position dependent friction coefficient. The posted equation, even with a null RHS, doesn't describe the system's behavior
So, how could you be sure of this result,do you have any suggestions on what to do or how to think?
 
  • #6
I think you don't have a result, yet.
Hints: think about energy and work. Where does the work needed to compress the spring come from?
 
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  • #7
Gordianus said:
I read the posted question and found the friction doesn't follow a simple, viscous-like, form. It states a position dependent friction coefficient. The posted equation, even with a null RHS, doesn't describe the system's behavior
I know that work is equals the change in the particle’s kinetic energy.Now, we have a particle with initial velocity and it has also kinetic energy.We have friction force oppsite direction of the motion and we have also spring force.Then I thought if I wrote down (1/2mvinitial^2+kx=1/2mvfinal^2+friction force )but I don't think this might be right.It does not look like it is correct.There is a problem but I can't figure out.
 
  • #8
You're adding apples (energy) to oranges (force).
Try again.
 
  • #9
phase0 said:
Oh,thank you!Here is my question,I just did not share the original version because I did not want to be understood as solving my question to you
Let’s see your free body diagram of the mass
 
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  • #10
I believe the problem does not give enough information to calculate the value asked by a).
 
  • #11
Lnewqban said:
I believe the problem does not give enough information to calculate the value asked by a).
I believe there is enough information. I wrote the differential equation and found ##v(x)## then applied the given condition that ##v(D)=0##. This is equivalent to using the work-energy theorem as others have suggested. What piece of information do you think is missing?
 
  • #12
kuruman said:
I believe there is enough information. I wrote the differential equation and found ##v(x)## then applied the given condition that ##v(D)=0##. This is equivalent to using the work-energy theorem as others have suggested. What piece of information do you think is missing?
I don’t get it, I am missing something for sure.
D point may or may not be reached, depending on how much kinetic energy is involved.
I will wait to see development of the solution.
 
  • #13
Lnewqban said:
D point may or may not be reached,
D is defined as the point reached.
 
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  • #14
haruspex said:
D is defined as the point reached.
Not quite. D is a displacement. According to the statement of the problem (post #3, figure 1.png), "The maximum compression of the spring is D."
 
  • #15
kuruman said:
Not quite. D is a displacement. According to the statement of the problem (post #3, figure 1.png), "The maximum compression of the spring is D."
But x is measured from the tip of the relaxed spring, so the displacement and coordinate are equal.
 
  • #16
haruspex said:
But x is measured from the tip of the relaxed spring, so the displacement and coordinate are equal.
Indeed it is. However, if the author's intention were "point D" as opposed to "displacement D" the statement would have been something like "At maximum compression, the mass is at x = D." We are mincing words here about the definition of "D" and 'tis not worth it. Instead, 'tis the season to be jolly - Happy holidays.
 
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  • #17
As I understand it, the problem says "At maximum compression, the mass is at x=D. Besides the strange friction coefficient, it's a nice example of the work-energy theorem.
 
  • #18
kuruman said:
We are mincing words here ... 'tis the season to be jolly
That's ok, I like mince pies.
 
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  • #19
phase0 said:
... do you have any suggestions on what to do or how to think?
As first noted by @Gordianus, think work/energy.

What expression gives the frictional force on the mass? Can you now derive an expression for the work done by the frictional force when the mass moves from x=0 to x=D? (Hope that's not too big a 'hint'.)

If you can do that, the rest should be straightforward.
 
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Related to Hooke's law does not apply when there is friction, right?

1. What is Hooke's law and how does it relate to friction?

Hooke's law states that the force exerted by a spring is directly proportional to the displacement of the spring from its equilibrium position. This law only applies when there is no friction present.

2. Why doesn't Hooke's law apply when there is friction?

Friction is a force that opposes motion and is caused by the interaction between two surfaces. When friction is present, it can absorb some of the force that would normally be exerted by the spring, resulting in a different relationship between force and displacement.

3. Can Hooke's law be applied to real-world situations where friction is present?

In most real-world situations, friction is present and therefore Hooke's law cannot be applied directly. However, it can still provide a good approximation in some cases, such as when the frictional force is relatively small compared to the force exerted by the spring.

4. Are there any other factors that can affect the applicability of Hooke's law?

Yes, there are other factors that can affect the applicability of Hooke's law, such as the material properties of the spring and the presence of any other external forces acting on the system.

5. Is there a way to account for friction when using Hooke's law?

Yes, there are ways to account for friction when using Hooke's law, such as by including a frictional force term in the equations or by using experimental data to determine the effect of friction on the system.

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