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Two blocks and a pulley homework

  1. Oct 25, 2008 #1
    1. The problem statement, all variables and given/known data

    The two blocks are connected by a light string that passes over a frictionless pulley with a negligible mass. The 3 kg block lies
    on a rough horizontal surface with a constant coefficient of kinetic friction 0.1. This block
    is connected to a spring with spring constant 5 N/m. The second block has a mass of 7 kg.
    The system is released from rest when the
    spring is unstretched, and the 7 kg block falls
    a distance h before it reaches the lowest point. acceleration of gravity is 9.8m/s/s
    Calculate the falling distance h where the
    7 kg blocks stops. Answer in units of m.
    Note: When the 7 kg block is at the lowest point velocity is zero

    and then
    calculate the moment when the 7kg block has descended by a distance of 10.5056m, where 10.5056 is less than h. at this moment calc the sum of kinetic energy for the two blocks in joules.


    2. Relevant equations
    E=mgh= M m work
    (1/2mv^+1/2kx) + (1/2mv^2+ mg(h-x) + (f sub k * x)
    this equaition was given by a friend but I can't seem to make it work
    h=2(m-uM)g/k
    any help on setting up the problem would be appreciated

    for the second part here's what I got thus far total k.e.= 1/2mv^2 +1/2Mv^2= mgx - 1/2kx^2-uMgx
    but I'm not sure about it.

    there's a picture of the diagram here

    [​IMG]
     
    Last edited: Oct 25, 2008
  2. jcsd
  3. Oct 26, 2008 #2

    Doc Al

    User Avatar

    Staff: Mentor

    The basic idea is energy conservation:
    Mechanical Energy(initial) + Work done by friction = Mechanical Energy(final)

    Mechanical energy is the sum of spring PE (½kx²), gravitational PE, and KE.
    Work done by friction (which is negative) = -μNx
     
  4. Oct 26, 2008 #3

    okay I'll try it out
     
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