Question

An engine lifts 400 kg mass through a height of
100 m in 20 s. The rated power of the engine is, if
efficiency of engine is 80% (g = 10 m/s2)
(1) 30 kW
(2) 20 kW
(3) 25 kW
(4) 10 kW​

Answer 1

Answer:

20kw

Explanation:

work done will be WD = mgh

here m=400kg , h=100m,g=10m/s^2

WD = 400×10×100

=4 × 10^5 or 4 ×10^2 × 10^3

=400kwh

power = work done / time

= 400kwh/20sec

= 20kw answer

Answer 2

Given:

Mass = 400kg

Height = 100m

Time = 20s

Engine efficiency = 80%

To Find:

The rated power of the engine

Solution:

Calculating input energy -

= 400 × g × H

Ei = 400 × 10 × 100

= 400000J

Energy per sec = Input energy/Time

= 400000/20

= 20000

= Converting into Kw

= 20000/1000

= 20kw

Answer: The rated power of the engine is 20kw