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The Work - Energy Theorem in a pulley?

  1. Sep 24, 2005 #1
    Consider the system shown in the figure. The rope and pulley have negligible mass, and the pulley is frictionless. The coefficient of kinetic friction between the 8.00-kg block and the tabletop is [tex]\mu_k=0.250.[/tex] The blocks are released from rest.

    Use energy methods to calculate the speed of the 6.00-kg block after it has descended 1.50 m.

    [tex]\Delta * K[/tex] = [tex](1/2)(m)(v_f^2 - v_i^2)[/tex]

    But, it was at rest, so [tex](1/2)(m)(v_f^2)[/tex] is what matters.

    What do I do next?

    Attached Files:

  2. jcsd
  3. Sep 24, 2005 #2


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    Homework Helper

    You will have a loss in PE balanced by a gain in KE and work done.

    What is it that loses PE, and by how much ?

    What are the gains in KE and by how much ?

    What work is done and how much ?
    Last edited: Sep 24, 2005
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