The Work - Energy Theorem in a pulley?

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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Fermat

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 ?