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%
