How Do Friction and Spring Constants Affect Block Velocities?

[chahud42]

Homework Statement

A light spring of force constant 3.90 N/m is compressed by 8.00 cm and held between a 0.250 kg block on the left and a 0.530 kg block on the right. Both blocks are at rest on a horizontal surface. The blocks are released simultaneously so that the spring tends to push them apart. Find the maximum velocity each block attains if the coefficient of kinetic friction between each block and the surface is the following. In each case, assume that the coefficient of static friction is greater than the coefficient of kinetic friction. Let the positive direction point to the right.

We have to find the final velocities of both masses for μ_k values of .000, .100, and .483.

Homework Equations

Conservation of momentum, Newtons second law, conservation of energy

The Attempt at a Solution

Initially, the blocks are at rest and the spring is compressed. So the initial momentum of the system is 0, but there is a force in the spring which is equal to F=kx=(3.9Nm)(.08m)=.312N
So, once the spring is released, it pushes on each block with this force. The momentum of each block is increased to P₁=m₁v₁ and P₂=m₂v₂. We can't go much further with this because the only 2 unknowns with using conservation of momentum are the velocities, which is what we need.

So I started to look at F=kx=ma. I can use kx/m=a to solve for the accelerations of the blocks caused by the spring. I got that a₁=1.25m/s² and a₂=.59m/s². Then I wanted to use this in the formula v²=v_o²+2aΔx to solve for the velocity. For the velocity of block one i got v=.45m/s. This was wrong so I didn't even bother to go further because something in my solution is wrong. None of my classmates or I can solve this. Where did we go wrong?

One thought I had was that the force exerted on each block won't be equal to kx, but kx/2 because the force is shared over two separate masses. But I tried this and this didn't work either. I'm fresh out of ideas.

 

[haruspex]

the formula v²=v_o²+2aΔx

That is only for constant acceleration.
For the zero friction case you could use conservation laws. For the others you might need to develop the differential equation, but start by considering static friction. You are told this is at least as great as the kinetic. Draw the FBD for each block.

We have to find the final velocities of both masses

No, you are asked for the maximum velocities. When the kinetic friction is nonzero the "final" velocities will all be zero.

 

[chahud42]

That is only for constant acceleration.
For the zero friction case you could use conservation laws. For the others you might need to develop the differential equation, but start by considering static friction. You are told this is at least as great as the kinetic. Draw the FBD for each block.

No, you are asked for the maximum velocities. When the kinetic friction is nonzero the "final" velocities will all be zero.

Right that's what I meant by saying final velocity since its at the maximum right before it loses contact with the spring. I'm still stuck on it. I get I can't use the constant acceleration case, but does that mean what I was doing before I plugged my numbers into that equation, what I had was right?

 

[chahud42]

That is only for constant acceleration.
For the zero friction case you could use conservation laws. For the others you might need to develop the differential equation, but start by considering static friction. You are told this is at least as great as the kinetic. Draw the FBD for each block.

No, you are asked for the maximum velocities. When the kinetic friction is nonzero the "final" velocities will all be zero.

Actually, I was just thinking, if the energy is conserved here (which I do believe it is) would it be possible to use the equation .5kx²=.5mv²? By using this you could solve for v and substitute that back in the conservation momentum equation right? My only problem is I don't understand which block has that velocity, or is it both?

 

[haruspex]

if the energy is conserved here

Only in the frictionless case.

which block has that velocity

They may have different velocities, so consider the total KE.

 

[chahud42]

Only in the frictionless case.

They may have different velocities, so consider the total KE.

Ok so it would be more like .5kx²=.5m₁v₁²+.5m₂v₂²? In the case with friction would you have to use the work energy theorem: ΔKE=W_spring(This I'm not sure about but I know that the spring is doing work on this blocks and the friction is taking away work) -W_f? If I use this, I wouldn't know how to incorporate the elastic potential energy from the spring because it only deals with a change in the kinetic energy. I feel like I have most of it but I'm missing important points. I get you're trying to lead me to the answer but I'm going to need a bit more. Assignment was already turned in anyway, I just still want to solve this.

 

[haruspex]

Ok so it would be more like .5kx²=.5m₁v₁²+.5m₂v₂²?

Yes.

In the case with friction

In the case with friction, remember that initially everything is stationary. So what is the first thing to check?

 

[chahud42]

Yes.

In the case with friction, remember that initially everything is stationary. So what is the first thing to check?

Yes.

In the case with friction, remember that initially everything is stationary. So what is the first thing to check?

Yes.

In the case with friction, remember that initially everything is stationary. So what is the first thing to check?

You have to check if the force of the spring overcomes the force of static friction first to see if it moves. So you have to analyze F_s=-kx and compare it to N=μmg for each block. If kx is bigger than μmg, then that means that the block will be accelerated. Once you do that though, I don't understand how to solve for the two separate velocities from .5kx²=.5m₁v₁²+.5m₂v₂². Could you deal with each block separately by saying .5kx²=.5m₁v₁² and .5kx²=.5m₂v₂². Does the spring exert the same force on each block or is the force shared between the two blocks?

 

[haruspex]

You have to check if the force of the spring overcomes the force of static friction first to see if it moves.

Right.

Once you do that though, I don't understand how to solve for the two separate velocities from .5kx²=.5m₁v₁²+.5m₂v₂².

You can only use that in the no friction case. How about you solve the frictionless case first, then we can turn to the frictional case?

 

[chahud42]

Right.

You can only use that in the no friction case. How about you solve the frictionless case first, then we can turn to the frictional case?

Conservation of momentum killed me in high school and now in college
Ok to solve it without friction, you can use .5kx²=.5m₁v₁²+.5m₂v₂² since energy is conserved, and 0=m₁v₁+m₂v₂ since momentum is also conserved. Solving for v₁ using momentum i get v₁=-m₂v₂/m₁. Substitute this expression for v₁ in the equation for conservation of energy and you get

. Solve for v₂ so that

. Plug in, get v₂, then plug all of those numbers in the original conservation of momentum equation and solve for v₁. Look right? I don't feel like bothering with numbers right now.

 

[haruspex]

Look right?

No, your algebra went awry when you substituted for v₁. The result was dimensionally wrong.
Take it in smaller steps.

 

[chahud42]

No, your algebra went awry when you substituted for v₁. The result was dimensionally wrong.
Take it in smaller steps.

Damn i can't even believe I left out a whole half of the equation. I swear I'm not usually this bad I should be writing it out. Its also Friday and i want to finish this problem I've been doing for 2 days hah. So REsubstituting (much slower now) v₁, i got

The algebra here seemed to get a bit tricky to keep control of but I double checked it, I think that's right.

 

[haruspex]

I think that's right.

So do I.

 

[chahud42]

So do I.

Great. Could you push me in the right direction for the cases with friction? Solving energy and momentum problems with friction is a pain because the force is non conservative. Is it possible to start with the same equation and account for friction doing negative work on each block?.5kx²=.5m₁v₁²-μ₁N₁d+.5m₂v₂²-μ₂N₂d

 

[haruspex]

Great. Could you push me in the right direction for the cases with friction? Solving energy and momentum problems with friction is a pain because the force is non conservative. Is it possible to start with the same equation and account for friction doing negative work on each block?.5kx²=.5m₁v₁²-μ₁N₁d+.5m₂v₂²-μ₂N₂d

You seem to have forgotten what you posted earlier:

You have to check if the force of the spring overcomes the force of static friction first to see if it moves.

Do that for both blocks in each of the nonzero friction cases.

 

[chahud42]

You seem to have forgotten what you posted earlier:

Do that for both blocks in each of the nonzero friction cases.

Well in the problem they didn't define a μ_s value for any parts. You can't just compare it with μ_k because that's lower than μ_s. I mean, it can tell you if the force can't overcome the kinetic friction, but it has to get past static first.

 

[haruspex]

Well in the problem they didn't define a μ_s value for any parts. You can't just compare it with μ_k because that's lower than μ_s. I mean, it can tell you if the force can't overcome the kinetic friction, but it has to get past static first.

Right, but remember

assume that the coefficient of static friction is greater than the coefficient of kinetic friction.

Does that resolve some cases at least?