Physics 220 Exercise on power
The net force exerted on the baggage by the machine is equal to and opposite the force exerted by the Earth: mg. The angle between the force vector F and the velocity vector v is 60° , so the power required to move the baggage is:
Now we compare with the power available from the power curve of the machine. The power required is a straight line curve with slope mg/2. The maximum bag mass corresponds to the steepest line we can draw and still intersect the power curve. Since the power curve itself is linear between v = 0 and v = 3 m/s, the steepest line lies along the power curve and has slope (10 kW)/(2 m/s) = 500 kW· s/m. This corresponds to baggage of mass:
Now we draw a line with slope mg/2 = (500 kg)(10 m/s2)/2 = 2500 kg· m/s2 and find where it intersects the power curve. The intersection is at the far right, and the speed is about 3.8 m/s.