The 500 kg video display will be raised to its position 20 m above the
arena floor by a motorized winch powered by a 12 volt battery. If the
battery capacity is 100 A-hr and the system is 100% efficient, how many
times could the display be lifted into position before the battery requires
recharging?
work to lift the video is 500*20 = 10,000 J
1 A-hr-V = 3600J
To determine the number of times the display can be lifted into position before the battery requires recharging, we need to calculate the total energy consumption for each lift and then divide the battery capacity by this value.
First, let's calculate the gravitational potential energy required to lift the display. The formula for gravitational potential energy is:
Potential energy = mass * gravitational acceleration * height
Given:
Mass of the display (m) = 500 kg
Height (h) = 20 m
Gravitational acceleration (g) = 9.8 m/s²
Potential energy = 500 kg * 9.8 m/s² * 20 m = 98,000 J
Since the system is 100% efficient, the electrical energy required would be the same as the potential energy calculated above.
To convert this energy into units of watt-hours (Wh), divide by 3600 (since 1 watt-hour = 3600 joules):
Energy (in watt-hours) = 98,000 J / 3600 = 27.22 Wh
Now we can calculate how many times the display can be lifted before the battery requires recharging. Divide the battery capacity by the energy consumption per lift:
Battery capacity = 12 volts * 100 A-hr = 1200 Wh
Number of lifts = Battery capacity / Energy consumption per lift
Number of lifts = 1200 Wh / 27.22 Wh ≈ 44 times
Therefore, the display can be lifted into position approximately 44 times before the battery requires recharging.