Consider the system shown in ?Fig. P6.81?. The rope and pulley have negligible mass, and the pulley is frictionless. Initially the 6.00-kg block is moving downward and the 8.00-kg block is moving to the right, both with a speed of 0.900 m/s. The blocks come to rest after moving 2.00 m. Use the work–energy theorem to calculate the coefficient of kinetic friction between the 8.00-kg block and the tabletop.

Solution 87P The forces on each block are According to work energy theorem, work done = change in kinetic energy W total (K.E) 2 (K.E) 1 (K.E) = m v + m v 1 2 1 2 A 1 2 B 1 (K.E) 2 0 The tension T on the rope does positive work on block B and the same magnitude of negative work on block A, so T does no net work on the system . Gravity...