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Spring and weight, acceleration

  1. Nov 5, 2014 #1

    rlc

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    1. The problem statement, all variables and given/known data
    A spring has an equilibrium length of 20.0 cm and a spring constant of 51.9 N/m. The spring is connected to the underside of the roof of a car and a 0.214 kg block suspended from it. How long (in cm) is the spring when the car is at rest?
    How long is the spring if the car is accelerating horizontally at 1.63 m/s2?

    2. Relevant equations
    mgh=0.5kh^2
    mg=0.5kh

    3. The attempt at a solution
    Nothing is working for me and I don't know what I'm missing!
     
  2. jcsd
  3. Nov 5, 2014 #2

    BvU

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    "Nothing is working for me" ? Did nothing sign some contract to work for you then ? :)

    You are supposed to know how a spring behaves when a weight is hung from it. What's the force gravity excerts on a block of mass m ? What's the extension of a spring with constant k when a force F is applied ?
    Set up a force balance: mg down, kx up. No acceleration, no net force: mg - kx = 0
     
  4. Nov 5, 2014 #3

    rlc

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    According to Hook's law, F=-kx
    where k=51.9 N/m
    and x is the displacement.
    Also, force=mg=0.214g*9.8=2.0972 N

    Any hints as to where to go next?
     
  5. Nov 5, 2014 #4

    BvU

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    You mean other than "Set up a force balance: mg down, kx up. No acceleration, no net force: mg - kx = 0" ?
     
  6. Nov 5, 2014 #5

    rlc

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    mg - kx = 0
    (2.0972 N)-(51.9N/m)(x)=0
    x=0.040408 m (the displacement)
     
  7. Nov 5, 2014 #6

    rlc

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    Ok, I got the first question:
    At rest with no acceleration, the x=0.040408 m is the displacement. Convert that to cm and Add the equilibrium length of 20 cm gives the length of the spring when car is at rest. (24.041 cm)
     
  8. Nov 5, 2014 #7

    rlc

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    But what about when the car accelerates at 1.63 m/s^2?
     
  9. Nov 6, 2014 #8

    BvU

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    Well, the block doesn't disappear through the rear window of the car, so something must be accelerating it to the tune of 1.63 m/s2. That something must exercise a net force in the horizontal direction. There is only one thing in the exercise that can do that.

    So now you want to extend your collection of relevant equations under 2) with an equation that features the acceleration, force and mass. Any candidate ?

    Make a few drawings: one of a mass hanging from a spring in the car at rest, one that represents your idea of a mass hanging from a spring in an acclerating car.
     
  10. Nov 6, 2014 #9

    rlc

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    ma=kx
    m(a+g)=kx
    0.214 kg(1.63+9.8)=51.9x
    0.047129m
    4.7129 cm+20=24.7129 cm

    But this is wrong. I messed up somewhere. What do you recommend doing?
     
  11. Nov 6, 2014 #10

    rlc

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    SQRT[(1.63)^s+(.214*9.8)^2]
    =2.65615 m/s^2=the total acceleration

    Is this on the right track?
     
  12. Nov 6, 2014 #11

    rlc

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    F=ma
    F=(0.214)(2.65615)
    F=0.5684 N
     
  13. Nov 6, 2014 #12

    BvU

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    Firing away again, eh ? Physics is a science, not a lucky shot affair !

    m(a+g)=kx is a bit strange. Net acceleration = m x net force. Net force = sum of forces, i.e mg - kx = ma. Not in numbers: all these forces are vectors.

    What I reccomend doing is the same as what I recommended doing: Make the drawing to see which points where.
     
  14. Nov 11, 2014 #13

    rlc

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    Free body diagram would be drawn with the mass hanging from the spring being the center point. From this, the Weight force downward (m*acceleration downward), and acceleration (I drew this going to the left), and the Tension of the spring (I drew this upward and towards the left) can be drawn.

    Force of spring cos(theta)=W

    From this equation we can get two new ones:
    K(delta L)sin(theta)=ma
    K(delta L)cos(theta)=mg

    If you divide the two equations, you get tan(theta)=a/g
    You can use this to calculate theta, and then use either of the two equations to calculate for (delta L)

    BUT, I did it a bit differently. You can square both of the equations to get rid of needing to know what theta is.
    (K^2)(delta L^2)=(m^2)(g^2+a^2) ..........watch the parenthesis here..........

    So with my numbers: (51.9^2)(delta L^2)=(0.214^2)(9.8^2+1.63^2)
    2693.61(delta L^2)=4.5199
    (delta L^2)=0.001678
    Delta L=0.040963 meters
    Multiply by 100 to convert to centimeters (4.0963)
    And add the equilibrium length of the spring (+20)
    My answer ended up being 24.10 cm
     
  15. Nov 11, 2014 #14

    BvU

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    I fully agree. The angle wrt horizontal is 80.5 degrees instead of 90, so that's quite noticeable, but the extension is only marginally different (0.556 mm) from the no-acceleration case (because of the vector addition of the forces).
     
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