Efficiency of Motor: 6V, 0.25A, 200g Weight, 4J Potential Energy

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The discussion revolves around calculating the efficiency of an electric motor used to lift a 200g weight through a height of 2m, resulting in a potential energy increase of 4J. The first experiment involved a voltage of 6V and a current of 0.25A, with the weight taking 5 seconds to rise. Participants clarified that efficiency is determined by the ratio of useful energy output to energy input, where output is based on the weight's height and mass, and input is derived from voltage, current, and time. Despite some students feeling unprepared due to a lack of prior instruction on efficiency, they expressed a desire to understand the calculation process. The conversation emphasizes the importance of grasping these fundamental concepts in physics.
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A student performs an experiment in which an electric motor is used to lift a 200g weight through 2m, thus increasing its potential energy 4J. from measurements of the rate at which the weight is lifted the efficiency of the motor is to be determined. Two different voltages were used and the current was measured.
a) in the first experiment at 6V a current of 0.25A was measured and the weight took 5s to rise the 2m what was the efficiency of the motor.

Ok. well in our class we haven't even learned about effiency so i have no idea how to approach, although our teacher said we don't need to know this type of Questions for exams or anything i would still like to know how to do it. Could anyone please explain thanks!
 
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usagi101 said:
… in the first experiment at 6V a current of 0.25A was measured and the weight took 5s to rise the 2m what was the efficiency of the motor.

Ok. well in our class we haven't even learned about effiency so i have no idea how to approach, although our teacher said we don't need to know this type of Questions for exams or anything i would still like to know how to do it. Could anyone please explain thanks!

Hi usagi101! :smile:

Efficiency just has its usual English meaning … amount of energy obtained divided by amount of energy supplied.

The energy obtained is calculated from the height and the mass.

The energy supplied is calculated from the volts the amps and the time.

Have a go! :smile:
 
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