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A mass hung from two attached springs

  1. Oct 27, 2011 #1
    1. The problem statement, all variables and given/known data

    A light spring with constant k1 is hung from an elevated support. From its lower end a second light spring is hung, which has spring constant k2. An object of mass m is hung at rest from the lower end of the second spring.

    a) Find the total extension distance of the pair of the springs.

    2. Relevant equations

    F= ma
    Fs = kx

    3. The attempt at a solution

    I treated the two springs as a single device and made the following equation for the forces acting on the mass:

    Fnet = 0
    Fs1 + Fs2 - mg = 0
    k1x + k2x = mg
    (k1 + k2)x = mg
    x = mg/(k1 + k2)

    But this answer is incorrect. (Correct answer: x = mg(1/k1 + 1/k2).

    Would someone be able to clarify what is happening in this system? I seem to be missing something...
     
  2. jcsd
  3. Oct 27, 2011 #2
    Seems like you need to treat each spring as a separate system. They have different spring constants so will put out different forces.

    k1x = mg gives x1=mg/k1 similarly x2 = mg/k2
    add x1 and x2 gives you total distance the spring is stretched.

    Just my $0.02, this is my first year taking calculus based physics as well.
     
  4. Oct 27, 2011 #3
    Hm I see. I was wrong to assume the forces on each spring would be the same...

    Thank you! And I hope your physics course(s) are going well :)
     
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