Question

An electric motor of power 200 w is switched on for 1 minute and 40 seconds. If 60 percent of the energy of the motor is useful,calulate (a) useful work done by the motor is

Answer 1

Given, power of electric motor , P = 200w
motor is switched on for 1minute and 40 sec
so, time , t = 60 + 40 = 100 sec.

we know, Energy = power × time
= 200 × 100 = 20000 J

A/C to question,
60% of the energy of the motor is useful.
so, useful energy = 60 % of 20000
= 60/100 × 20000
= 12000 J

hence, useful workdone by the motor = 12000J

Answer 2

Given that,

time = 100 seconds

power = 200 watt

energy = power X time

= 200 X 100 = 20000 joule

among this energy,

only 60% is utilised

therefore,

energy useful for work done by the motor is,

60/100 X 20000

= 12000 joule