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Kinetic Energy Methods

  1. Apr 30, 2007 #1
    1. The problem statement, all variables and given/known data

    Consider the system shown. The coefficient of kinetic friction between the 8.00 kg block and the tabletop is ųk = 0.25. Neglect the mass of the rope and of the pulley, and assume that the pulley is frictionless. Use energy methods to calculate the speed of the 6.00 kg block after it has descended 2.00 m., starting from rest.


    2. Relevant equations

    i have no idea. help me.

    3. The attempt at a solution

    i have no idea. help me.
  2. jcsd
  3. Apr 30, 2007 #2
    someone help me please...
  4. Apr 30, 2007 #3
    welcome to PF. Do you have any ideas? Hint: the kinetic energy of the two blocks comes from the potential energy of the falling block.
    So, if there were no friction:
    mgh=1/2(m+M)v^2 where m=6Kg, and M=8Kg but there is friction, do you know how to compute work done by friction?

    Normally we ask for some thoughts or at least posting some eqns on a problem, before anyone will help.
  5. May 1, 2007 #4
    this is what i did:
    first i solved for the potential energy
    PE = mgh = 6(9.8)(2) = 117.6 J

    then i solved for the force done on the block
    F = ų*mg = (0.25)(6)(9.8) = 14.715 N

    then i solved for the work done on the block
    W = (14.715)(2) = 29.43

    then i solved for the kinetic energy
    KE = PE - W = 117.6 - 29.43 = 88.17 J

    and lastly i solved for the speed
    V = (88.17/0.5(6))^-2 = 5.42 m/s

    am i right?
    Last edited: May 1, 2007
  6. May 1, 2007 #5
    Your on the right track
    [tex]$ U_p=2*6*9.8=117.6 J$[/tex]
    But you've used the wrong mass in the friction force
    [tex]$ F_f=0.25*8*9.8=19.6N\\
    W=2*F_f=39.2 J\\
    Your final line also contains a mistake as you have assumed that the 6kg blosk is the only object with kinetic energy, whereas both object have the same velocity.
    [tex]$ E_k=\frac{1}{2}(m+M)v^2=7v^2\\
    7v^2=78.4 J\\
  7. May 1, 2007 #6
    thanks a lot!
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