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Spring Connected between 2 Masses

  1. Mar 7, 2014 #1
    1. The problem statement, all variables and given/known data

    A light spring of force constant 4.00 N/m is compressed by 8.00 cm and held between a 0.250 kg block on the left and a 0.560 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.

    uk = 0.000, 0.100, 0.431

    k = 4.00 n/m
    x = 8 cm
    m1 = 0.250 kg
    m2 = 0.560 kg

    2. Relevant equations

    1/2 kx^2
    1/2 mv^2
    W =F*D


    3. The attempt at a solution

    1/2(4.00)*(0.08) = 1/2*(.56)*(v)^2
    v = -0.08*sqrt(.32/.56) = -0.604 m/s

    I'm having major issues with this problem :( if possible please explain all the math you did to figure out each final velocity and explain how to set it up :(

    I know in the case that the larger m2 with uk = 0.100 is 0 m/s and the last uk value makes both masses 0 m/s
     
    Last edited: Mar 7, 2014
  2. jcsd
  3. Mar 7, 2014 #2

    BvU

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    Hello MDP and (a late) welcome to PF,

    Let's start with ##f = 0##. You will be familiar with action = - reaction and there are no external horizontal forces in the problem. Are you also familiar with momentum conservation?

    Since there are no external horizontal forces, momentum of center of mass is conserved:
    $$\vec F_{1\rightarrow 2} = - \vec F_{2\rightarrow 1} \Leftrightarrow {d\vec p_1\over dt} = {-d\vec p_2 \over dt} \Leftrightarrow {d\over dt} \left ( \vec p_1 + \vec p_2 \right ) = 0$$

    That means both masses move apart with equal momentum (otherwise the c.o.m. would not stay in place). No friction means all spring energy is converted to kinetic energy.

    Enough info to get started. With friction it becomes a little more complicated, but you should be able to do the first one. Show us and then we move on to the second. The third one (I guess) leaves one of the blocks in place.

    [edit] Oh, I see you already saw that 2 leaves one block in place and 3 leaves both in situ. How did you conclude that ?
     
  4. Mar 7, 2014 #3
    my homework has a pratice version and it said those values were zero i kinda guessed because i tried setting it up having Wext + Usi + k1i + k2i = k1f + k2f + Usf

    but i got a nasty equation in the end i can't solve for vf

    looks like this (keep in mind I just started calculus this semester so we can use algebra for this course)

    initial energy + Wexternal = final energy

    Wext + Usi + k1o + k2o = k1f + k2f + Usf

    uk*m*g*delta x1-uk*m2*delta x2-m1iv1i = m2v2

    delta x1 = -m2/m1 delta x2
    delta x2 = -m1/m2 delta x1

    v2 = -m1v1/m2

    after much manipulation of kinetic energy and the formula above i got this for a final

    v2^2 = -4*uk*g*x2 + kxo^2/m2 - k/m2 (xo-x2(m1/m2+1))^2

    all that divided by m2/m1 + 1
     
  5. Mar 7, 2014 #4
    my professor gave me

    1/2kx^2 = 1/2m2v2f^2 + 1/2m1v1f^2

    oh and the xo = 8 cm and the k = spring constant for the long formula in previous post
     
  6. Mar 7, 2014 #5

    BvU

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    says spring energy = total kinetic energy. Two unknowns: the speeds. One other equation: conservation of momentum. Done!
     
  7. Mar 7, 2014 #6
    how would i use the conservation of momentum to solve for the velocities?
     
    Last edited: Mar 7, 2014
  8. Mar 7, 2014 #7
    The problem statement doesn't tell you what the value of the coefficient of static friction is. All it tells you is that it is higher than the coefficient of kinetic friction. So how do you know that the blocks aren't glued to the table, in which case the coefficient of static friction would be infinite, and the blocks would never move? This is a very poorly worded problem statement.

    Chet
     
  9. Mar 7, 2014 #8
    yes i blame my book for it. so mathematically how would I set up the momentum to solve for both velocities?
     
  10. Mar 7, 2014 #9
    Do you have to use momentum balance approach, or are you allowed to set up the problem in plain old Newton's second law force balance? If the latter, then here's one way of doing it:

    Let m1=0.25 kg and m2=0.56 kg
    Let x1 be the x coordinate of m1 and x2 be the x coordinate of m2.
    Let L be the original unloaded length of the spring.

    In terms of x1, x2, and L, how much has the spring been stretched?
    In terms of x1, x2, L, and k, what is the tension in the spring?
    If the spring is compressed 0.08 m initially, what is the relationship between the initial locations of the two masses x10, x20, and L at time t = 0?
    In terms of x10, x20, L, and k, what is the initial tension in the spring?

    We are going to do a force balance on each of the two masses, and we are going to include in the force balances the unit vector in the x-direction [itex]\vec{i}[/itex]. If T is the tension in the spring, do a force balance on m1. Here is the force balance for the case in which there is no friction:

    [tex]m_1\frac{d^2x_1}{dt^2}\vec{i}=T\vec{i}[/tex]

    What is the corresponding force balance for m2?

    We'll continue after you've provided some answers to these questions.

    chet
     
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