IIT JEE , AIEEE Forum - Physics , Mechanics

Xavier4

It is desired to lift 2000 kg of metal

through a distance of 12 m in 1

minute.Find the minimum horsepower

of the engine to be used?

Decoder

work done by lift = - work done by gravity..

(assuming tht kinetic energy doen't change during lifting..)

so W = 2000 x 12 x 10...

so power = 2000 x 12 x 10 / 60..Watts..

this gets u 4kW..

so one horsepower is 746 or 776..do the conversion now..!!

sarthak321

see work has to be done against gravity

taking PE 0 a ground
and using work energy theorem

work(force) = mgh + 1/2mv^2

but for minimum work v=0

work = 2000 X 10 X 12
= 240000 J

now power = 240000/60

= 4000 W
since 1 hp = 746 W

4000 W = 4000/746 hp

rate me if the answer helped

reddevil_2009

Power=Work done/Time teken
          = Force.Displacement/Time Taken
          = 2000*10*12/60
          =4*10^3 W
746 W=1hp
4*10^3 W=4*10^3/746

Just calculations left now
Plzzzz rate me......

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