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What is the stretch of the spring?

  1. Jul 26, 2015 #1

    KAC

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    1. The problem statement, all variables and given/known data
    A spring scale hung from the ceiling stretches by 6.3 cm when a 1.2 kg mass is hung from it. The 1.2 kg mass is removed and replaced with a 1.4 kg mass. What is the stretch of the spring?

    F=mg
    U (spring force) = 1/2kx^2

    2. Relevant equations
    The units of cm need to be converted into m, so 6.3cm becomes 0.063m. Units of kg are kept the same.
    To solve the problem we need to find k, the spring constant.

    3. The attempt at a solution
    First, I'm going to use the first situation with the displacement and mass given to determine the spring constant.

    F = mg; F = (1.2kg)(9.8m/s^2); F = 11.76N
    F = U;
    U = 1/2kx^2
    F = 1/2kx^2
    k = 2F/x^2
    k = 2(11.76N)/0.063m^2
    k = 5925.9

    From there I plugged the spring constant into the second situation.

    F = mg; F = (1.4kg)(9.8m/s^2) = 13.72N
    U = 1/2kx^2
    x = sqrt of 2*F/k
    x = sqrt 2(13.72N)/5925.9
    x = 0.068 m or 6.8 cm.

    This final answer for displacement is incorrect and I am unsure of my mistake. Did I make a calculation error or set the problem up incorrectly?

    Thank you in advance for your help!
     
  2. jcsd
  3. Jul 26, 2015 #2
    Where did you get that U = F from?

    Chet
     
  4. Jul 26, 2015 #3

    KAC

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    Chet,

    I used U = F as the force; the equation for U that I used is the potential energy of the spring. Since it is being pulled down and there is no normal force in this case, only the weight force is acting on the object which is = mg. (mass*gravity).
     
  5. Jul 26, 2015 #4

    jbriggs444

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    Science Advisor

    Spring force varies linearly with displacement. Spring energy varies as the square of the displacement. Did you choose the wrong equation?
     
  6. Jul 26, 2015 #5

    KAC

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    Two other formulas I may be able to use are Force of the spring = -kx (where x is displacement) and L(change in length) = mg/k. But both of these equations have two variables I do not know in them. When you say that spring energy varies as the square of the displacement, this leads me to believe that U = 1/2kx2 is the correct equation since it is varying as the square of whatever the displacement is. But since the only other force acting on the mass is it's weight force since it is oscillating vertically, I am not quite understanding why it is incorrect to use mg = 1/2kx2. Is it incorrect because the weight is moving constantly (oscillating) and isn't in a set position? So therefore the weight force does not always equal the spring force?
     
  7. Jul 26, 2015 #6

    jbriggs444

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    Science Advisor

    What are the units of the spring constant? What are the units of mg? What are the units of 1/2 kx2?
     
  8. Jul 26, 2015 #7

    KAC

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    The spring constant, k, has units of N/m; mg = F which is N (N = kg*m/s2), and U = N/m * m = N as well since it is the specific force in this case. So setting F = U in this case should be acceptable since the units are the same and U is the specific force that is working on the object, yes?
     
  9. Jul 26, 2015 #8
    What would you say if I told you that the units of U are Nm? All energy has units of Nm=Joules. You must have learned this previously, correct?

    Chet
     
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