Friction, energies and variable distance

In summary: So, just focus on the energy at the first spring. Use the energy method exactly as you started. (But I'm assuming this is not the way the problem was intended to be solved.)In summary, the problem involves a rail with two springs of equal constant K and length D on each side, and an object of mass M with a coefficient of friction μ. The object is given an initial potential energy of KD^2/2 by forcing it against one of the springs until it is fully contracted and then releasing it. The question is where will the object stop, assuming infinite static friction and that it does not start moving again once it stops. The solution involves finding the work of friction on the distance x = 0 to
  • #1
Chen
977
1
Suppose you have a rail with two springs on the sides, both with the same constant K. The length of the rail is L, and the length of each spring is D (when not contracted or elongated). On that rail you have an object of mass M (which may be treated as a point particle), and the coefficient of friction between that object and the rail is μ. You take the object, and give it an initial potential energy of KD2/2, by forcing it against one of the springs until it is fully contracted and releasing the object.

Where will the object be brought to a halt? Assume that once it stops, it does not start moving again (i.e infinite static friction).

Obviously the beginning of the answer should be:
[tex]F_{friction}x = \Delta {E_p}_{els}[/tex]
But where do we go from there? We don't know where the object stops, so we don't know if it still has potential energy. Furthermore, how do we find the final location of the object from x?
 
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  • #2
First, I assume that the length dimension in the problem is L+2D,
that is L is the length of that part of the rail which lies strictly between the unstretched springs (?).
Secondly, as to your infinite static friction:
Where do you want it to occur?
If you exclude it from the starting point, it still will make the system stop terribly soon.
Thirdly, since you say you give the object an initial potential energy of K/2D^(2), this should be the same as saying the velocities of the object and the tip of spring remains equal up to x=D (if starting at x=0); the particle then parts from the spring and is no longer imparted any energy from the stretching of the spring.
In particular, the particle's potential energy at D is 0.

Without having more detailed information of how the spring is constructed, I don't see how we otherwise might solve the problem.
In particular, I assume that a spring is always unstretched when the particle is not in contact with it.

We can readily calculate the work of friction on the distance x=0 to D,
and hence V(D,t1).
(If imaginary, the object stopped somewhere x<D)

Along the rail, we may calculate the velocity loss, and find V(D+L,T1),
just prior to production of new potential energy.

Hence, the contraction C1 of the second spring should be given by:
1/2mV(D+L,T1)^(2)=1/2KC1^(2)+(mu)*mgC1
, where (mu)*mgC1 is the frictional work.
With infinite static friction, this gives your answer.
 
  • #3
No, the total length of the rail is L, including the unstretched springs. Forget about the static friction too, let's just say we are looking to find where the object first stops. What else do you need to know?
 
  • #4
The problem is a bit harder if you only have a finite static friction;
the stopping criterion (when a spring is stretched) then becomes K*Cn<=(mustat)*mg,
where (mustat) is the coefficient of static friction, and Cn the n'th spring contraction...
 
  • #5
1. Velocity at x=D:
V(D,t1)=sqrt(((K/M)D^(2)-2(mu)*gD)
If nonzero and real:
2. On the strecth x=D, L-D, only frictional force (that's what I've assumed), hence:
1/2MV1(L-D.T1)^(2)=1/2MV1(D,t1)^(2)-(mu)*Mg(L-2D)
If nonzero and real:
3.
1/2MV1(L-D.T1)^(2)=1/2KC1^(2)+(mu)*MgC1.

Position of particle should then be L-D+C1.
 
  • #6
Chen said:
Obviously the beginning of the answer should be:
[tex]F_{friction}x = \Delta {E_p}_{els}[/tex]
But where do we go from there? We don't know where the object stops, so we don't know if it still has potential energy. Furthermore, how do we find the final location of the object from x?
I'm not getting the difficulty of this problem. (I'm probably missing something.)

First find out if the mass stops before the spring fully decompresses, that is at x < L. (Ignore the fact that infinite static friction would not allow the mass to start moving. :rolleyes: ) Use the energy method that you started. If the mass does stop before x = L, then you are done. If not, you have some remaining KE at x = L. Then find out if it stops before hitting the second spring. Etc, etc.
 
  • #7
The problems is that the object is allowed, and does, bounce between the two springs several times before stopping. So x is most definitely larger than L.
 
  • #8
Chen said:
The problems is that the object is allowed, and does, bounce between the two springs several times before stopping. So x is most definitely larger than L.
Ah... that just means that your values for k, μ, L, and D are such to allow it. Since the contraints are different in the three regions (x = 0-D, D-(L-D), (L-D)-L), I doubt there's a simple algebraic way to write down the answer. But given a particular set of parameters, I'm sure you'd have no problem cranking out the answer. (But I bet you knew that! :smile: )
 
  • #9
I thought I should try finding the energy that the object loses in every cycle, and from there find the difference in the spring's contraction. The problem is that the energy loss depends on the spring's contraction, so it's like a loop... I will try doing it 'manually', one cycle at a time, and see where that gets me.
 
  • #10
On second thought, given your "infinite static friction" contraint, the mass will never make it to x = L. If it has the energy to make it to the second spring, that spring will surely stop it.
 

1. What is friction?

Friction is a force that resists the motion of two objects in contact with each other. It is caused by the microscopic roughness of surfaces and can be affected by factors such as the type of material and the force pressing the objects together.

2. How does friction affect energy?

Friction converts kinetic energy (energy of motion) into thermal energy (heat), which means that some of the energy used to move an object is lost as heat due to friction. This can lead to a decrease in the overall energy of a system.

3. What factors affect friction?

The factors that affect friction include the type of surfaces in contact, the force pressing the surfaces together, the roughness of the surfaces, and the speed at which the objects are moving. Additionally, the presence of lubricants or surface coatings can also affect friction.

4. How does distance affect friction?

The distance between two objects can affect the amount of friction between them. The closer the objects are, the greater the surface area of contact and therefore the greater the friction. However, if the distance between the objects is too far, there may not be enough force to create significant friction.

5. How can friction be reduced?

Friction can be reduced by using lubricants such as oil or grease, which create a layer between two surfaces and decrease the amount of direct contact. Additionally, using smoother or more slippery materials, such as Teflon, can also reduce friction. Reducing the force pressing two surfaces together can also decrease friction.

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