Including kinetic friction into Work of spring

[Ambidext]

Homework Statement

2 masses, are attached to a spring as follows: (trying my best to illustrate)

[M1]----[M2] (the ---- represents a spring between 2 masses)

M1 = 0.25kg
M2 = 0.5kg

The spring is compressed 0.08m, and then the 2 masses are released simultaneously. If the kinetic friction coefficient between the masses and the floor(both masses are of same material) is 0.1 and the spring has a spring constant of 3.85N/m, what is the maximum speed M1 and M2 can reach upon release?

Homework Equations

f_k = 0.1 mg
W_s = 0.5 k x²

The Attempt at a Solution

I can calculate the speeds if there's no friction. The system can be viewed as a system of linear momentum, and that the energy of the spring is being transferred to 2 masses, initially at rest, and M1v1 + M2v2 = 0

After which, I can use conservation of kinetic energy to solve for v1 and v2. But with friction, which is a force, I do not know how to include it in these equations with the absence of a time variable. Any guidance would be appreciated!

 

[Ambidext]

Anyone? :(

 

[PhanthomJay]

The velocity is maximum when the acceleration is zero. This follows from the definition of acceleration, a = dV/dt, where 'a' is the slope (tangent to the curve at any point) of the V vs. t curve. When a is zero, the veocity is at a maximum or minimum value. Without friction, this maximum velocity occurs at the equilibrium point (unstretched spring point) at its first return to that position , the value of which can be found through conservation of energy methods. When friction is involved, you first have to determine at what point the acceleration is zero. One way to do this is to examine forces and apply Newton's law that F_net = 0 at the point of no acceleration.

 

[Ambidext]

I'm sorry, but I am still unclear at the direction of your approach...

 

[PhanthomJay]

Well, sure, it is not an easy problem. Because of friction, an external force, momentum is not conserved, so you need to take a different approach. You first need to accept the fact that the maximum speed of the blocks occur when the acceleration is 0, which I tried to explain in my first response. So you have to determine the point where a = 0, which, per Newton's first law, is the point where all the forces (the net force) acting on each block is zero. Can you identify that point? You can find it in terms of an unknown variable, x, by drawing free body diagrams of each block, and identifying the forces actiong on each of them in the horizontal direction (Hint, friction and spring forces act in the horizontal direction, which must be equal and opposite to each other for the equilibrium condition of F_net = 0). Then use conservation of energy principles, (including the work done by friction), or use the work energy theorem if you wish (but which usually bogs you down with the plus and minus signs.)

 

[Ambidext]

Well, sure, it is not an easy problem. Because of friction, an external force, momentum is not conserved, so you need to take a different approach. You first need to accept the fact that the maximum speed of the blocks occur when the acceleration is 0, which I tried to explain in my first response. So you have to determine the point where a = 0, which, per Newton's first law, is the point where all the forces (the net force) acting on each block is zero. Can you identify that point? You can find it in terms of an unknown variable, x, by drawing free body diagrams of each block, and identifying the forces actiong on each of them in the horizontal direction (Hint, friction and spring forces act in the horizontal direction, which must be equal and opposite to each other for the equilibrium condition of F_net = 0). Then use conservation of energy principles, (including the work done by friction), or use the work energy theorem if you wish (but which usually bogs you down with the plus and minus signs.)

Okay. So that also means, at maximum velocity, F_net = F_spring - f_k = 0, ie, f_k = F_spring?

If that is true, it should lead to:
F_spring = f_kinetic
3.85 (x) = 0.1 (M1+M2) g
x = 0.191

Does that even make sense?

 

[diazona]

This is a tricky problem indeed I would actually advise you to consider this: for the given initial conditions (spring compressed 0.08m etc.), is the force exerted by the spring even enough to overcome the force of friction on the larger block? What about the smaller block?

 

[Ambidext]

This is a tricky problem indeed I would actually advise you to consider this: for the given initial conditions (spring compressed 0.08m etc.), is the force exerted by the spring even enough to overcome the force of friction on the larger block? What about the smaller block?

I believe so. The question gives coefficient of static friction, which means the blocks are moving.

I haven't solved the question by the way. I don't think x = 0.191 makes sense at all ;(

 

[PhanthomJay]

I believe so. The question gives coefficient of static you mean kinetic friction, which means the blocks are moving.

I haven't solved the question by the way. I don't think x = 0.191 makes sense at all ;(

Just because the kinetic friction coefficient is given (and you may assume that the static friction coefficient is the same, or more), that doesn't mean that both blocks are moving. They both might be moving, or both might be not moving, or one might move and the other not move. Answer Diazona's questions again. At the instant the masses are released, what is the spring force acting on the larger block(draw a free body diagram of the larger block)? Is that enough to overcome the friction force acting on the larger block to make it move? Then look at the forces acting on the smaller block...is the spring force greater than the friction force acting on it at the moment of release? Does the smaller block move?

 

[Ambidext]

Just because the kinetic friction coefficient is given (and you may assume that the static friction coefficient is the same, or more), that doesn't mean that both blocks are moving. They both might be moving, or both might be not moving, or one might move and the other not move. Answer Diazona's questions again. At the instant the masses are released, what is the spring force acting on the larger block(draw a free body diagram of the larger block)? Is that enough to overcome the friction force acting on the larger block to make it move? Then look at the forces acting on the smaller block...is the spring force greater than the friction force acting on it at the moment of release? Does the smaller block move?

Hmm, okay. The force exerted by the spring should be:
F = kx = 3.85(0.08) = 0.308N

Does the spring exert 0.308N on EACH block, or is the TOTAL force acting on both blocks 0.308N?

EDIT:
After thinking awhile and drawing the free body diagram, it should be 0.308N EACH block correct? That means the smaller block would move but the bigger one wouldn't. Hope I'm right this time...

 

[PhanthomJay]

Hmm, okay. The force exerted by the spring should be:
F = kx = 3.85(0.08) = 0.308N

Does the spring exert 0.308N on EACH block, or is the TOTAL force acting on both blocks 0.308N?

EDIT:
After thinking awhile and drawing the free body diagram, it should be 0.308N EACH block correct?

Yes, initially at release, that is correct.

That means the smaller block would move but the bigger one wouldn't. Hope I'm right this time...

That also seems right, although the problem did not list the coefficient of static friction, which, if high enough, would prevent the smaller block motion as well. So one can only assume that the static and kinetic friction coefficients are equal or about equal.

 

[Ambidext]

Alright, thanks for the hints! :)