The electric motor of an elevator(lift) uses 630kJ of electric energy when raising the elevator and passengers, of total mass 12
,500N, through a vertical height of 29m.Calculate the efficiency of the elevator. pppppplease need helpp willl markk the brrainliest
1 answer:
Answer:
Efficiency = 0.575 = 57.5%
Explanation:
First, we will calculate the output energy of the elevator, which is equal to the potential energy acquired by the load.
![O = Potential\ Energy\\O = mgh](https://tex.z-dn.net/?f=O%20%3D%20Potential%5C%20Energy%5C%5CO%20%3D%20mgh)
where,
O = Output = ?
mg = W = Weight = 12500 N
h = height = 29 m
Therefore,
![O = (12500\ N)(29\ m) \\O = 362500\ J = 362.5\ KJ](https://tex.z-dn.net/?f=O%20%3D%20%2812500%5C%20N%29%2829%5C%20m%29%20%5C%5CO%20%3D%20362500%5C%20J%20%3D%20362.5%5C%20KJ)
The input is given as:
I = 630 KJ
Thus the efficiency will be:
![Efficiency = \frac{O}{I} = \frac{362.5\ KJ}{630\ KJ}\\\\](https://tex.z-dn.net/?f=Efficiency%20%3D%20%5Cfrac%7BO%7D%7BI%7D%20%3D%20%5Cfrac%7B362.5%5C%20KJ%7D%7B630%5C%20KJ%7D%5C%5C%5C%5C)
<u>Efficiency = 0.575 = 57.5%</u>
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