Asked by Allie
A 1.00 x 10^4 W electric motor is used to lift a 955 kg object vertically 25.0 m at a constant velocity in 45.0 s. What is the efficiency of the motor?
I did P=W/t (1.00x10^4 x 45 =W)
And got 4.50 x 10^5
Then I did
Ep' = Ek
955 x 9.81 x 25
= 234213.75
But I wasn't sure what these values meant, but I plugged them into the efficiency equation anyways
234213.75 / 450000 x100 = 52.0 % efficient.
I am just unsure of what the numbers really mean and what I am actually solving here.
I did P=W/t (1.00x10^4 x 45 =W)
And got 4.50 x 10^5
Then I did
Ep' = Ek
955 x 9.81 x 25
= 234213.75
But I wasn't sure what these values meant, but I plugged them into the efficiency equation anyways
234213.75 / 450000 x100 = 52.0 % efficient.
I am just unsure of what the numbers really mean and what I am actually solving here.
Answers
Answered by
bobpursley
Never do calculations if you are unsure of what they mean.
eff= power out/power in
power out= 955*g*25
power in= 1E4*45
caclulate efficiency. It is 52 percent. My experiences with electric motor (smaller than this) is that they are about 30 percent efficienct.
eff= power out/power in
power out= 955*g*25
power in= 1E4*45
caclulate efficiency. It is 52 percent. My experiences with electric motor (smaller than this) is that they are about 30 percent efficienct.
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