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 does work W mg = m Ad on block A, Block B moves horizontally ,so no work is done on it. Friction does work W = m gd on block B. fric k B Thus W total W mg + W fric = m Ad- m k B W total (K.E) 2 (K.E) 1 1 2 so m Ad mkg B (m +2m )A B 1 2 m A 1(mA+m B(0.9) k= m B+ 2 mBgd 6 kg (6 kg+8 kg)v k= + 1 2 1 8 kg 2 8 kg ×9.8 m/s ×2 m = 0.786