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Homework Help: Hooke's Law and suspended objects

  1. Jun 24, 2009 #1
    1. The problem statement, all variables and given/known data

    A body of mass (m) is suspended from a spring with spring constant k in configuration (a) and the spring is stretched 0.1m. If two identical bodies of mass (m/2) are suspended from a spring with the same spring constant k in configuration (b), how much will the spring stretch? Explain your answer.

    2. Relevant equations

    F = -kx (Hooke's Law)

    3. The attempt at a solution

    Since in the first case the mass is at rest,
    by mg - k(0.1) = 0, i calculated that k = 10mg

    In the second case, since the bodies suspended have a mass of (m[tex]\div[/tex]2), their weight is halved too. Therefore the spring will stretch for 0.1[tex]\div[/tex]2 = 0.05m on each side.

    i.e. It stretches for 0.1m (0.05m each on left and right)

    However, according to the solution, the answer is 0.05m.
    It is also stated that "if you have problems, substitute one of the two masses with a fixed hook; does the spring change its length?"

    What difference does that make?

    Attached is the diagram describing the setup

    Attached Files:

  2. jcsd
  3. Jun 24, 2009 #2

    Doc Al

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    Staff: Mentor

    Hint: The stretch of the spring is proportional to the tension. What's the tension in the spring in each scenario?
  4. Jun 24, 2009 #3
    Tension in the spring?

    Case 1:
    T = mg

    Case 2:
    T = mg/2
    But it is pulled from both sides!
    Do both blocks of objects with mass (m/2) cause the spring to stretch?

    This is getting confusing...
  5. Jun 24, 2009 #4

    Doc Al

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    Staff: Mentor

    It better be! Hint: The spring is being pulled from both sides in both scenarios.

    Sure. You need both blocks pulling to stretch the spring. Remove one, and the spring will go flying as the mass falls.
  6. Jun 25, 2009 #5

    In case 2:

    One of the masses is responsible for preventing the spring from flying out,
    while the other one causes the stretch?

    Therefore stretch length = 0.1/2 m = 0.05 m (since weight causing the stretch is halved)
  7. Jun 25, 2009 #6

    Doc Al

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    Staff: Mentor

    Don't think of it that way. To stretch a (massless) spring you must pull it from both ends with the same force--that force equals the tension in the spring.

    In case 1, the ceiling pulls up on one end while the mass pulls down on the other; both ends of the spring are pulled with a force equal to mg. The tension in the spring equals mg; it stretches accordingly.

    In case 2, each end is pulled with a force equal to mg/2. The tension in the spring equals mg/2; it stretches accordingly.
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