Electric Motor Efficiency: Maths Homework Answers

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Discussion Overview

The discussion revolves around a homework problem involving the efficiency of an electric motor used to lift a weight. Participants explore the calculations needed to determine the motor's efficiency based on given voltages, currents, and time taken for the weight to rise. The scope includes mathematical reasoning and conceptual understanding of efficiency in the context of physics experiments.

Discussion Character

  • Homework-related
  • Mathematical reasoning

Main Points Raised

  • A student presents an experiment with an electric motor lifting a 200g weight and seeks to calculate the efficiency based on provided data.
  • Some participants suggest that if the student's results are correct, they must have grasped some concepts correctly, encouraging them to ask specific questions for clarification.
  • One participant emphasizes the need to calculate the energy output of the motor at 6V and 0.25A over 5 seconds to find the efficiency, proposing a formula for efficiency as a percentage.
  • There is confusion among participants regarding the reference answers provided by the student, with some expressing a lack of understanding about them.

Areas of Agreement / Disagreement

Participants generally agree on the need to calculate energy output to determine efficiency, but there is confusion regarding the reference answers and the clarity of the initial problem statement. No consensus is reached on the interpretation of the provided answers.

Contextual Notes

Some participants note the importance of understanding the energy calculations involved in determining efficiency, but there are unresolved questions about the reference answers and their implications for the problem.

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Homework Statement



A Student Performs an experiment in which an electric motor is used to lift a 200g weight through 2m, thus increasing its potential energy by 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 an the current was measured.




Homework Equations



A/

In the first experiment at 6V, a current of 0.25A was measured and the weight took 5 seconds to rise the 2 metres. What was the efficiency of the motor?

B/

In the Next experiment the voltage was increased to 8V, the current was found to be 0.30A and the efficiency worked out to be 60%. How long did the motor take to lift the weight the 2m this time?


The Attempt at a Solution




I have no idea even how to work out efficiency yet so my attempts were quite pathetic


Answers for reference

a) 53%


b) 2.8 seconds
 
Last edited:
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Considering that your results are correct you must have understood something correct. Feel free to ask specific questions if you are unsure what exactly it is you have understood.
 
filiplarsen said:
Considering that your results are correct you must have understood something correct. Feel free to ask specific questions if you are unsure what exactly it is you have understood.

there are no results there, maaaaan.

Anyway, you should find the energy given off by the motor at 6V with 0.25 A for 5 s. (multiply the power with time and you should get energy). You will get a value in J. Efficency is found by Energy needed/Energy spent, and for expressing it in % you just multiply the result by 100.

So Eff= 100*4J/xJ %

Have fun!
 
Lok said:
there are no results there, maaaaan.

Ah, "answers for reference ". I did not get that.
 
filiplarsen said:
Ah, "answers for reference ". I did not get that.

Somebody had to put you on the right path ... :P
 

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