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Two Masses on Vertical Spring

  1. Oct 15, 2008 #1
    [​IMG]

    A block of mass 10 kg hangs on a spring. When a second block with an identical mass of 10 kg is tied to the first, the spring stretches an additional ho = 1.3 m.

    a) What is the value of the spring constant k?

    Now the string is burned and the second block falls off.

    b) How far above its original position does the remaining block attain its maximum speed?

    c) What is the maximum speed attained by the remaining block?




    Spring constant: F=kx
    Conservation of Mechanical Energy




    I was thinking about setting up the problem so you look at the first scenario with the 10kg block as zero potential energy. Then when the second 10kg block is added potential energy is gained...is this in the right direction at all?

    Could the change in mechanical energy in the first scenario equal the second?

    So...
    (1/2)mv(final)^2 + (1/2)kx(final)^2 = (1/2)mv(initial)^2 + (1/2)kx(initial)^2
    but that would just get rid of the k's which is what i'm looking for.
     
  2. jcsd
  3. Oct 15, 2008 #2
    I think i figured out part b, but i'm still stuck on part a.

    For b would you use the equation
    (1/2)kx^2 = (1/2)mv(final)^2 - (1/2)mv(initial)^2
    Knowing the spring constant and the distance its stretched all your left with unknown is the final velocity since the initial velocity is 0.

    But that would just give me the speed at the end of the distance stretched so that's not exactly right.
     
  4. Oct 15, 2008 #3
    alright i figured out part a,
    the weight of the added block equals the force of the spring
    10(9.81) = k (1.3m)
    k= 75.46N/m

    part b kind of has me scratching my head.

    So if the max speed occurs when the potential energy is zero how do i set that up?
     
  5. Oct 15, 2008 #4
    PEinitial + KEinitial = PEfinal + KEfinal
    PEinitial = KEfinal

    Right?
     
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