Question

A machine uses 1000 J of electric energy to raise a heavy mass, increasing its potential energy by 300 J. What is the efficiency of this process?

Answer 1

30 %

Step-by-step explanation:

Here machine uses 1000 J means input of machine is 1000 J

And its increases the potential energy by 300 J so output of the machine is 300 J

The efficiency of any system is defined as the ratio of output and input

So efficiency will be
= 30% so the efficiency of the machine will be 30%

Answer 2

Efficiency of this process is 30 %.

Step-by-step explanation:

It is given that,

Electric energy used by the machine,
`W_(in)=1000\ J`

The potential energy of the machine is increased,
`W_(out)=300\ J`

The efficiency of this machine is given by :

`\eta=(W_(out))/(W_(in))* 100`

`\eta=(300)/(1000)* 100`

`eta=30\%`

So, the efficiency of this process is 30 %. Hence, this is the required solution.