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

[usagi101]

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!

 

[tiny-tim]

… 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!

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!

 

[baba89]

As a scientist, it is important to understand the concept of efficiency and how it relates to motors and energy. Efficiency is a measure of how well a system can convert one form of energy into another. In this case, we are interested in the efficiency of the motor in converting electrical energy into potential energy of the weight.

To determine the efficiency of the motor, we can use the formula:

Efficiency = (Output energy/Input energy) x 100%

In this case, the output energy is the potential energy gained by the weight, which is 4J. The input energy is the electrical energy supplied to the motor, which can be calculated using the formula:

Input energy = Voltage x Current x Time

In the first experiment, the voltage was 6V and the current was 0.25A. The time taken for the weight to rise 2m was 5 seconds. Therefore, the input energy can be calculated as:

Input energy = 6V x 0.25A x 5s = 7.5J

Now, we can plug these values into the efficiency formula:

Efficiency = (4J/7.5J) x 100% = 53.3%

Therefore, the efficiency of the motor in the first experiment was 53.3%.

To determine the efficiency in the second experiment, we would need to know the values of voltage, current, and time. Once we have those values, we can follow the same steps as above to calculate the efficiency.

In conclusion, efficiency is an important concept in understanding the performance of motors and other energy conversion systems. By using the appropriate formulas and measurements, we can determine the efficiency of a motor and make improvements to increase its efficiency.