Answer:
91.9 %
Explanation:
The percentage efficiency is equal to the ratio between the output work and the input work:
\(\eta = \frac{W_{out}}{W_{in}} \cdot 100\)
The output work is equal to the gravitational potential energy gained by the refrigerator, which is equal to its weight (W=1,127 N) times the gain in height (h=0.85 m):
\(W_{out}=Wh=(1127 N)(0.85 m)=958 J\)
The input work is equal to the product between the force applied on the refrigerator (F=496 N) and the displacement of the refrigerator (d=2.1 m):
\(W_{in}=Fd=(496 N)(2.1 m)=1042 J\)
So, the efficiency is
\(\eta=\frac{958 J}{1042 J}\cdot 100=91.9 \%\)
