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/s^{2})/2 = 2500 kg· m/s^{2} and find where it intersects the power curve. The intersection is at the far right, and the speed is about 3.8 m/s.