An electric motor is used to lift an elevator and its load (total mass 1500kg) to a height of 20m. The time taken for the job is 20s. What work is done ? What is the rate at which work is done. If efficiency of the motor is `75%`, at which rate is the energy supplied to the motor?

Explanation

Here, `m=1500kg, h=20m, t=20s.`
Work done, `W=mgh=1500xx9.8xx20`
`=2.94xx10^(5)J`
Rate of doing work `= ("work")/("time")=(2.94xx10^(5))/(20)`
`=1.47xx10^(4)W`
As `eta=("output power")/("input power")`
`:.` Input power (= rate at which energy is supplied )
`=(1.47xx10^(4))/(75%)=(1.47xx10^(4)xx100)/(75)`
`=1.96xx10^(4)Js^(-1)`