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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 8.00kg 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 6.00kg 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?
Use energy methods to calculate the speed of the 6.00kg 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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