Question

A 3.5 kW drill transfers 5 000 kJ of kinetic energy during 15 seconds of use. What is the percentage efficiency of the drill?

Answer 1

1.04%

Step-by-step explanation:

Given that,

The power of drill = 3.5 kW = 3500 W

Transferred kinetic energy = 5000 kJ during 15 seconds of use.

We need to find the percentage efficiency of the drill. It can be given by :

`\eta=(P_o)/(P_i)* 100`

Where

Po and Pi are output and input powers.

`P_o=(5000* 10^3)/(15)\\\\=3.34* 10^5\ W`

So,

`\eta=(3500)/(3.34*10^(5))*100\\\\=1.04\%`

So, the percentage efficiency of the drill is 1.04%.