Question

Calculate the power of an engine required to lift 10^5 Kg of Coal per hour from a mine 360m deep, take g as 10 m/s^2.


plz don't post irrelevant answer​

Answer 1

m=10

5

kg,g=10ms

−2

,h=360m,t=1h=3600s

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

And, power, P=

t

W

=

t

mgh

∴P=

3600

10

5

×10×360

W=10

5

W=100kW

Answer 2

given,

m = 10⁵ kg , g = 10 ms -² , h = 360m t= 1hr = 3600s

w = mg × h

Power = w/t = mgh / t

p = 10⁵ × 10 × 360 / 3600 w

= 10⁵ w = 100kW