Question

An engine lifts a body 500 kg weight up to a height of 20 m in 10 sec. Find:
(1) the total work done by the engine.( ans : 10^5 joule)
(2)the power of the engine in Kilowatt.(ans:10 kilowatt)

Answer 1

hello mate here is ur answer

, given : (i)m=500kg , h=20m and t=10s

since we know that
W= F×S
w= m×g×s. (since object is lifted)
w= 500×9.8×20 = 98,000 joule or nearly equal to 1,00,000 joule if u take gravity equal to 10 instead 9.8

so hence, work done is 10^5 joule

(ii) power

since we know that

power = work done÷ time

therefore==

p= 10^5÷10

p= 10^4 watt ==(1)

but they have mentioned that it should be in kilo watt so hence divide the equation by 10^3
(since kilo=10^3)

that implies=
p= 10^4÷10^3

p=10 kilo watt

so hence here are ur answers mate

hope it helped u

#####be brainly######$$#

Answer 2

1) the total work done by the engine

w = mgh
= 500×20×10
= 100000
= 10^5 joule ans



2) the power of the engine in Kilowatt

= P = energy/unit time = mgh/t

= 500×20×10/10
= 10^4 watt
= 10 kilowatt ans

I hope it help you