erik-the-red
Sep24-05, 04:00 PM
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 \mu_k=0.250. 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.
\Delta * K = (1/2)(m)(v_f^2 - v_i^2)
But, it was at rest, so (1/2)(m)(v_f^2) is what matters.
What do I do next?
Use energy methods to calculate the speed of the 6.00-kg block after it has descended 1.50 m.
\Delta * K = (1/2)(m)(v_f^2 - v_i^2)
But, it was at rest, so (1/2)(m)(v_f^2) is what matters.
What do I do next?