Given data,
Initial velocity u = 0
Final velocity v = 6m/sec
Distance traveled s = 5m
For finding acceleration, using the relation, v2 = 42 + 2as
∴ a = 3.6 m/sec2 62 = 0 + 2a X 5 ... (i)
Apply the relation F = ma,
Let T = Tension in the string, same for both side
Using the relation F = ma, for block A
T – 100 = ma T – 100 = (100/9.81) X 3.6 ...(ii)
Using the relation F = ma, for block B
100 + F – T = ma
100 + F – T = (100/9.81) X 3.6 ...(iii)
Add equation (2) and (3), we get
F = 2[(100/9.81) X 3.6]
F = 73.5N.