12 . a = 5.0 m/s^2
T = 15 N13 . a = 3.3 m/s^2
T 1 = 13 NT 2 = 20 N14 . a = 0.22 m/s^2
T 1 = 20 NT 2 = 29 NStep-by-Step Solution to #2:
Step 1 : Free-body diagrams:
No components are necessary, so on to Step 3 : write Newton’s second law for each block, calling the
rightward direction positive:
Step 4 : Solve algebraically. It’s easiest to add these equations together, because the tensions cancel:
F = (3m )a , so a = F /3m = (10 N)/3(1 kg) = 3.3 m/s^2 .To get the tension, just plug back into T − 0 = ma to find T = F /3 = 3.3 N.
Step-by-Step Solution to #5:
Step 1 : Free-body diagrams: