1. A motor winds the rope in B with a constant acceleration a. Because of the pulley system, the mass m in A slides in this way to the left over a horizontal axis. Between m and the left end point of the horizontal axis, there is a spring (constant k, unstretched length lo). The coefficient of kinetic friction between m and the axis is k. When SA equals 1.5 m, VB equals 6 m/s. For this situation, determine at that instant the velocity of A and the acceleration of A. Data: aB = 2 m/s, m = 2 kg, k = 150 N/m, lo = 2 m, = 0.3 2 m B S There are two ropes here to consider: the rope indicated in red, and the one in blue. For the red rope: let Sc and SB be the position of the smaller pulley and the position of the big pulley respectively, both starting from the left wall. The length of the red rope is than equal to SB + 2 (SB - Sc). From this, we find vc = 1.5 VB and ac = 1.5 aB. For the blue rope: Compare this question to exercise 12.215. This exercise is solved in full (for other values) in the movie A of exercise session 3 (on Toledo). Combination of these equations leads to the solution: VA is 15 m/s to the left, and a = 101 m/s to the left.