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Dynamics Question (Kinetics& energy)

  1. Jun 10, 2010 #1
    1. The problem statement, all variables and given/known data

    The two blocks are connected by a light inextensible cord, which passes around small

    massless pulleys as shown below. If block B is pulled down 500 mm from the equilibrium

    position and released from rest, determine its speed when it returns to the equilibrium



    2. Relevant equations

    T1 + U1 = T2 + U2

    3. The attempt at a solution

    If it's pulled down below equilibrium and held there, then T1 of the system is zero because both blocks are not moving. At the moment B passes through the equilibrium, there is no more potential energy, only kinetic, then the equation would look like:

    U1 = T2

    The problem I'm running into is that I get a negative value for the left side of this equation, which is impossible because then it would have to go under a square root when solving for the velocity.

    for U1 I had:

    mgha - mghb + 0.5kx^2

    Since 'b' moves down 0.5m, a moves up 0.25 and the spring is stretched 0.25. Is this right since A is attached to the pulley and B is simply hanging? If that's the case then:

    (2)(9.81)(0.25) - (10)(9.81)(0.5) + 0.5(800)(0.25)^2

    Which is negative. Supposedly the answer is 2.16 m/s.
  2. jcsd
  3. Jun 10, 2010 #2


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    Science Advisor
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    Gold Member

    You neglected the fact that the spring is already stretched in the equilibrium position before the block B is pulled down.
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