The efficiency of an electric motor is 20 %. When lifting a body 500 J of energy are wasted. What is the useful work done by the motor?
please and thank u!
Eff. = Eo/Ei = 0.2.
Eff. = Eo/(Eo+500) = 0.2.
Eo = 0.2Eo+100,
0.8Eo = 100,
Eo = 125J. = Energy out = Useful work.
sir, the answer is supposed to be 125?? is there something wrong
HENRY THANK YOU MAN
Thanks a lot, Terrific information.
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Well, well, well! It seems like our electric motor needs a little boost in efficiency. Let's see what we can do here. If the efficiency of the motor is 20%, that means 20% of the energy is being put to good use, while 80% is just taking a vacation from productivity.
Given that 500 J of energy is being wasted, we can calculate the total energy input by dividing it by the efficiency. So, let's get to it:
Total energy input = Energy wasted / Efficiency
Total energy input = 500 J / 0.20
Now, if we subtract the wasted energy from the total energy input, we'll find the useful work done by the motor:
Useful work done = Total energy input - Energy wasted
I'll leave you to perform the final calculation, my friend. But remember, with a little efficiency improvement, that electric motor will be running like a well-oiled clown car!
To find the useful work done by the motor, we need to first understand the concept of efficiency. Efficiency is defined as the ratio of useful work output to the total energy input. In this case, the efficiency of the electric motor is given as 20%, which means that 20% of the input energy is converted into useful work.
To calculate the useful work done by the motor, we can use the formula:
Useful work = Efficiency * Total energy input
Given that the efficiency is 20% and the body wasted 500 J of energy, we need to determine the total energy input first.
Total energy input = Useful work + Wasted energy
Substituting the given values, we have:
Total energy input = Useful work + 500 J
Now, substituting the value of the efficiency (20%) into the formula, we can calculate the useful work:
Useful work = Efficiency * Total energy input
Useful work = 0.2 * (Useful work + 500 J)
Simplifying the equation:
Useful work = 0.2 * Useful work + 0.2 * 500 J
Useful work = 0.2 * Useful work + 100 J
Bringing the terms containing the useful work to one side of the equation:
Useful work - 0.2 * Useful work = 100 J
0.8 * Useful work = 100 J
Dividing both sides of the equation by 0.8:
Useful work = 100 J / 0.8
Useful work = 125 J
Therefore, the useful work done by the motor is 125 J.
1/5 Effort=500
Effort=2500J
useful work=2500-500=???