Question

Calculate the power of an engine required to lift
105 kg of coal per hour from a mine 360 m deep.
( Take g = 10 m s-2).

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Answer 1

Answer:

Given, m= 10^5 kg, g= 10m/s², h= 360m,

t1 = 1h = 60×60 seconds = 3600 seconds.

The work needed in lifting a mass m to a height h against a force due to gravity is W=mgh

and,

power p = work done/time taken = mgh/t

p= 10^5 × 10 × 360 / 3600 = 10^5 W = 100kW.

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