The pulley system below uses a gasoline engine to raise a drill head up through a smooth drill pipe. The engine provides a constant force of 900 N. The drill head has a mass of 80 kg. Find the power that is being consumed by the motor (provided by the gasoline) at the instant the drill head has been raised 5 m., starting from rest. The engine has an efficiency of 0.65.

NB: The diagram of the pulley system is not shown but the information provided is sufficient to answer the question

Power = 2702.56 W

Explanation:

Let the power consumed be P

Energy expended = E = mgh

height, h = 5 m

E = 80 * 9.8 * 5

E = 3920 J

To calculate the time, t

From F = ma

F = 900 N

900 = 80 a

a = 900/80

a = 11.25 m/s²

From the equation of motion,

The drill head starts from rest, u = 0 m/s

Power, P = E/t

P = 3920/0.0.943

P = 4157.79 W

But Efficiency, E = 0.65

P = 0.65 * 4157.79

Power = 2702.56 W

the duration of a photographic flash is related to an rc time constant, which is 0.100 μs for a certain camera.