The power developed by the electric motor is equal to the work done per unit time.
The work done by the motor in lifting the load is given by:
W = mgh
where m is the mass of the load, g is the acceleration due to gravity, and h is the height through which the load is lifted.
Since the load is lifted at a constant speed, the height through which it is lifted is given by:
h = v^2/2g
where v is the speed of the load.
Substituting the given values, we get:
h = (3 m/s)^2 / (2 x 10 m/s^2) = 0.45 m
The work done by the motor is therefore:
W = mgh = 100 kg x 10 m/s^2 x 0.45 m = 450 J
The power developed by the motor is given by:
P = W/t
where t is the time taken to lift the load. Since the speed is constant, the time taken is given by:
t = h/v = 0.45 m / 3 m/s = 0.15 s
Substituting the values, we get:
P = 450 J / 0.15 s = 3000 W
Therefore, the electric motor develops a power of 3000 W.
The answer is option (a) 3000W.