wassayhassan07
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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

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hamzaahmeds

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.

[tex]O = Potential\ Energy\\O = mgh[/tex]

where,

O = Output = ?

mg = W = Weight = 12500 N

h = height = 29 m

Therefore,

[tex]O = (12500\ N)(29\ m) \\O = 362500\ J = 362.5\ KJ[/tex]

The input is given as:

I = 630 KJ

Thus the efficiency will be:

[tex]Efficiency = \frac{O}{I} = \frac{362.5\ KJ}{630\ KJ}\\\\[/tex]

Efficiency = 0.575 = 57.5%

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