Answer:
A power of 29562.1 W is required to operate the tow.
Explanation:
Given:
Angle of inclination (x) = 15.0°
Length of slope (L) = 300 m
Mass of each rider = 70 kg
Number of riders = 50
Speed of rope (v) = 12 km/h
Now, mass of the 50 riders (m) = Mass of 1 rider × Number of riders
= 70 × 50 = 3500 kg
The force acting on the tow is the component of weight of 50 riders in the direction opposite to the motion of the tow.
The component of weight along the slope is given as:
[tex]F_{slope}=mg\sin x=3500\times 9.8\times \sin 15^\circ = 8877.5\ N[/tex]
Now, the tow must apply the same force in the direction of motion to move it at constant speed.
Now, speed is in km/h. So, we need to convert it to m/s using the conversion factor.
We know that, 1 km/h = 5/18 m/s
Therefore, 12.0 km/h = [tex]12\times \frac{5}{18}=3.33\ m/s[/tex]
Therefore, the minimum power required to operate the tow is given by the formula:
Power = Force × Velocity
[tex]P=8877.5\times 3.33=29562.1\ W[/tex]
Therefore, a power of 29562.1 W is required to operate the tow.
