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Hooke's Law: Find x (How much will the spring stretch)?

  1. Apr 8, 2015 #1
    1. The problem statement, all variables and given/known data
    | Mass 1: 2kg | -------- | Mass 2: 3kg | ----------> 15N
    2 blocks are connected by a massless spring on a horizontal frictionless plane. The left block has a mass of 2kg and the right block has a mass of 3 kg. There is a force pulling Mass 2 to the right with a force of 15N.
    Spring Constant = 130N/m

    2. Relevant equations
    Fspring = -kx

    3. The attempt at a solution
    So I subbed in k into the equation.
    Fspring = -130x
    At this point I don't know what to do next. I've tried subbing 15N for Fspring but then I'll get a negative x value.
    Am I missing another equation that I'm supposed to use or a value that I'm supposed to find?
     
  2. jcsd
  3. Apr 8, 2015 #2
    Let T be the tension in the spring (forget about T = kx for now). Draw a free body diagram on Mass 1, showing the forces acting on it. Draw a free body diagram on Mass 2, showing the forces acting on it. Write a force balance equation on Mass 1. Write a force balance equation on Mass 2. Show us what you get so far.

    Chet
     
  4. Apr 8, 2015 #3
    So what I drew is basically both blocks having a downward force of mg and an upward force of N of equal magnitude as mg.
    For Mass 1, I drew an arrow to the right as T (tension).
    For Mass 2, I drew an arrow to the right as F (15N pulling the mass to he right) and an arrow to the left as T (tension of the spring pulling Mass 1)

    For writing a force balance equation, sorry to disappoint, but I don't know where to start with that.
    But I'm going to guess T (from Mass 1) = T (from Mass 2)
     
  5. Apr 8, 2015 #4
    Good job.

    You basically have it. For each mass, now set the net force on it equal to the mass times its acceleration.

    Chet
     
  6. Apr 8, 2015 #5
    Mass 1:
    Fnet = ma
    15 = 2a
    a = 15/2
    a = 7.5m/s2

    Mass 2:
    Fnet = ma
    15 = 3a
    a = 15/3
    a = 5m/s2
     
  7. Apr 8, 2015 #6
    This is not correct. It is not consistent with the forces you identified in your previous post as acting on each of the masses.

    This is what I was looking for:

    ##M_1a=T##

    ##M_2a=15-T##

    (If the amount of stretch in the spring is constant, then the two masses have to be accelerating with the same acceleration a.)

    What do you get if you add the above two equations together?

    Chet
     
  8. Apr 8, 2015 #7
    T = 15-T
    2T = 15
    T = 15/2
    T = 7.5N
     
  9. Apr 8, 2015 #8
    No. Your algebra is not correct. Adding the two equations together gives:

    ##(M_1+M_2)a=15##

    ##a=3 m/s^2##

    Substitute this value for a into either of the force balance equations to get T. What do you obtain? What does that give for the amount that the spring is stretched?

    Chet
     
  10. Apr 8, 2015 #9
    :nb) Sorry, argh, shouldn't get something so simple wrong! :nb)

    So I'm guessing T = Fnet
    Fnet(m1) = ma
    = 2*3
    = 6N

    Fnet(m2) = ma
    = 3*3
    = 9N

    T = kx
    Since Mass 2's forces are balanced (guessing here), I should use Fnet(m1).
    6 = kx
    6 = 130x
    x = 6/130
    x = 0.046 metres (3dp)

    Cheers Chet :smile:
     
    Last edited: Apr 8, 2015
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