Given data,

Initial velocity u = 0

Final velocity v = 6m/sec

Distance traveled s = 5m

For finding acceleration, using the relation, v^{2} = 4^{2} + 2as

∴ a = 3.6 m/sec^{2} 6^{2} = 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.**