- #1
curlydafatboy
- 5
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a motor furnishes 120 hp (746W=1hp) to a hoisting device that lifts 5000kg load to height of 13.0 meters in a time of 20.0 seconds. Find the efficiency...please help sum1 at least a formula?
The efficiency of lifting a 5000kg load 13m in 20 seconds can be calculated by dividing the work done (W) by the input energy (E) and multiplying by 100%. This calculation would look like: Efficiency = (W/E) x 100%. In this case, the work done would be 5000kg x 9.8m/s^2 x 13m = 637,000 Joules. The input energy would be the power (P) times the time (t), which would be 2000 watts x 20 seconds = 40,000 Joules. Plugging these values into the formula, the efficiency would be (637,000/40,000) x 100%, which equals 1592.5%.
The efficiency of motor lifting a load will typically decrease as the weight of the load increases. This is because heavier loads require more work to be done, which in turn requires more energy input. As a result, the efficiency of the motor will decrease as it needs to work harder to lift a heavier load.
There are several factors that can affect the efficiency of motor lifting, such as the weight and height of the load, the power and type of motor, the condition of the motor, and the use of any pulleys or gears. Other external factors like friction and air resistance can also impact the efficiency of motor lifting.
The efficiency of motor lifting can be improved by using a more powerful motor, reducing the weight of the load, and minimizing any external factors that may cause resistance. Additionally, regular maintenance and proper lubrication of the motor can help improve its efficiency.
Yes, there is a theoretical maximum efficiency that can be achieved when lifting a load with a motor. This is known as the Carnot efficiency and is dependent on the temperature difference between the input and output energy. In practical terms, the efficiency of motor lifting will always be less than this theoretical limit due to factors like friction and energy loss.