Answer:
30 %
Explanation:
The efficiency of a machine is defined as:
[tex]\epsilon = \frac{E_{useful}}{E_{input}}[/tex]
where
[tex]E_{useful}[/tex] is the useful output energy
[tex]E_{input}[/tex] is the energy in input
In this problem,
- The useful output energy is the increase in potential energy of the mass, so 300 J
- The energy in input to the machine is 1000 J
Therefore, the machine's efficiency is
[tex]\epsilon = \frac{300 J}{1000 J}=0.30[/tex]
In percentage,
0.30 x 100 = 30 %
