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.
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0
Answer:
100kW
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
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