2.09 m/s
OK, you have 2 masses.. m1 = 5.40 kg and m2 is 80% of that or 0.80 * 5.40 = 4.32 kg.
With the setup, The total mass of the system is
5.40 kg + 4.32 kg = 9.72 kg
The kinetic gained from the falling mass is
5.40 kg * 2 m * 9.8 m/s^2 = 105.84 kg m^2/s^2
The kinetic energy lost from the rising mass is
4.32 kg * 2 m * 9.8 m/s^2 = 84.672 kg m^2/s^2
So the total kinetic energy gained by the system is
105.84 kg m^2/s^2 - 84.672 kg m^2/s^2 = 21.168 kg m^2/s^2
Now this energy has been working on a system with a total mass of 9.72 kg, so
21.168 kg m^2/s^2 / 9.72 kg = 2.177777778 m^2/s^2
over a distance of 2m, so
2.177777778 m^2/s^2 / 2 m = 1.088888889 m/s^2
So the effective acceleration your system is undergoing is
1.089 m/s^2
Now to calculate the speed. Distance under constant acceleration is d = 0.5 A T^2, so plugging in the known values and solving for T gives:
2 m= 0.5 * 1.089 m/s^2 T^2
2 m = 0.5445 m/s^2 T^2
3.673094582 s^2 = T^2
1.92 s = T
Multiply the time by the acceleration gives:
1.92 s * 1.089 m/s^2 = 2.09 m/s
Note, the acceleration of 1.0888889 m/s^2 can be easily obtained using the relative masses of the two objects. We have an object of mass 1 and a second object of mass 0.8 for a total mass of 1.8, this mass is being accelerated by the difference of 0.2, so the effective acceleration is 0.2/1.8 = 0.111111 times that of the local gravity, which is 9.8 m/s^2. And 0.11111 * 9.8 = 1.088889 m/s^2.
