1. Sep 24, 2005 #1

   erik-the-red

   

   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?

    

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   erik-the-red, Sep 24, 2005
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3. Sep 24, 2005 #2

   Fermat

   

   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

   Fermat, Sep 24, 2005

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