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

Answer:

Efficiency of this process is 30 %.

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 :

$\eta=\dfrac{W_{out}}{W_{in}}\times 100$

$\eta=\dfrac{300}{1000}\times 100$    

$eta=30\%$

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

Answer 2

Answer

30 %

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 $\eta =\frac{output}{input}=\frac{300}{1000}=0.3$ = 30% so the efficiency of the machine will be 30%