Question

A refrigerator with a weight of 1,127 newtons needs to be moved into a house using a ramp. The length of the ramp is 2.1 meters, and its height is 0.85 meters. What is the percentage efficiency if a force of 496 newtons is applied to the refrigerator?

Answer 1

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

$\eta=\frac{958 J}{1042 J}\cdot 100=91.9 \%$

Answer 2

The used work of energy output is 496*2.1=1041.6 J. And the actual work of energy output is 1127*0.85=957.95 J. The percentage efficiency is 957.95/1041.6*100%=91.97 %.