In the diagram, M = 6 kg and m = 4 kg. Assume that the tabletop is frictionless. Determine the acceleration of the blocks once they are released from rest. (Round to the nearest tenth digit)
The tension of the cord is the same everywhere on the cord.
The accelerations of the two masses are the same.
T = ma
Mg T = Ma
Mg ma = Ma
a = Mg ÷ (m + M)
a = (6 × 9.8) ÷ (4 + 6)
a = 5.9 m/s2